Articles

4.5: Triple Integrals


Learning Objectives

  • Recognize when a function of three variables is integrable over a rectangular box.
  • Evaluate a triple integral by expressing it as an iterated integral.
  • Recognize when a function of three variables is integrable over a closed and bounded region.
  • Simplify a calculation by changing the order of integration of a triple integral.
  • Calculate the average value of a function of three variables.

Previously, we discussed the double integral of a function (f(x,y)) of two variables over a rectangular region in the plane. In this section we define the triple integral of a function (f(x,y,z)) of three variables over a rectangular solid box in space, (mathbb{R}^3). Later in this section we extend the definition to more general regions in (mathbb{R}^3).

Integrable Functions of Three Variables

We can define a rectangular box (B) in (mathbb{R}^3) as

[B = ig{(x,y,z),|,a leq x leq b, , c leq y leq d, , e leq z leq f ig}.]

We follow a similar procedure to what we did in previously. We divide the interval ([a,b]) into (l) subintervals ([x_{i-1},x_i]) of equal length (Delta x) with

[Delta x = dfrac{x_i - x_{i-1}}{l},]

divide the interval ([c,d]) into (m) subintervals ([y_{i-1}, y_i]) of equal length (Delta y) with

[Delta y = dfrac{y_j - y_{j-1}}{m},]

and divide the interval ([e,f]) into (n) subintervals ([z_{i-1},z_i]) of equal length (Delta z) with

[Delta z = dfrac{z_k - z_{k-1}}{n}]

Then the rectangular box (B) is subdivided into (lmn) subboxes:

[B_{ijk} = [x_{i-1}, x_i] imes [y_{i-1}, y_i] imes [z_{i-1},z_i],]

as shown in Figure (PageIndex{1}).

For each (i, , j,) and (k), consider a sample point ((x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)) in each sub-box (B_{ijk}). We see that its volume is (Delta V = Delta x Delta y Delta z). Form the triple Riemann sum

[sum_{i=1}^l sum_{j=1}^m sum_{k=1}^n f ( x_{ijk}^*, y_{ijk}^*, z_{ijk}^*),Delta x Delta y Delta z.]

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

Definition: The triple integral

The triple integral of a function (f(x,y,z)) over a rectangular box (B) is defined as

[lim_{l,m,n ightarrowinfty} sum_{i=1}^l sum_{j=1}^m sum_{k=1}^n f ( x_{ijk}^*, y_{ijk}^*, z_{ijk}^*),Delta x Delta y Delta z = iiint_B f(x,y,z) ,dV] if this limit exists.

When the triple integral exists on (B) the function (f(x,y,z)) is said to be integrable on (B). Also, the triple integral exists if (f(x,y,z)) is continuous on (B). Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, (f) is bounded on (B) and continuous except possibly on the boundary of (B). The sample point ((x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)) can be any point in the rectangular sub-box (B_{ijk}) and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s theorem for triple integrals exists.

Fubini’s Theorem for Triple Integrals

If (f(x,y,z)) is continuous on a rectangular box (B = [a,b] imes [c,d] imes [e,f]), then

[iint_B f(x,y,z) ,dV = int_e^f int_c^d int_a^b f(x,y,z) ,dx , dy , dz.]

This integral is also equal to any of the other five possible orderings for the iterated triple integral.

For (a, b, c, d, e) and (f) real numbers, the iterated triple integral can be expressed in six different orderings:

[egin{align} int_e^f int_c^d int_a^b f(x,y,z), dx , dy , dz = int_e^f left( int_c^d left( int_a^b f(x,y,z) ,dx ight) dy ight) dz = int_c^d left( int_e^f left( int_a^b f(x,y,z) ,dx ight)dz ight) dy = int_a^b left( int_e^f left( int_c^d f(x,y,z) ,dy ight)dz ight) dx = int_e^f left( int_a^b left( int_c^d f(x,y,z) ,dy ight) dx ight) dz = int_c^d left( int_a^b left( int_c^d f(x,y,z) ,dz ight)dx ight) dy = int_a^b left( int_c^d left( int_e^f f(x,y,z) ,dz ight) dy ight) dx end{align}]

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Example (PageIndex{1}): Evaluating a Triple Integral

Evaluate the triple integral [int_{z=0}^{z=1} int_{y=2}^{y=4} int_{x=-1}^{x=5} (x + yz^2), dx , dy , dz. onumber ]

Solution

The order of integration is specified in the problem, so integrate with respect to (x) first, then y, and then (z).

[egin{align*} int_{z=0}^{z=1} int_{y=2}^{y=4} int_{x=-1}^{x=5} (x + yz^2) , dx , dy , dz = int_{z=0}^{z=1} int_{y=2}^{y=4} left. left[ dfrac{x^2}{2} + xyz^2 ight|_{x=-1}^{x=5} ight], dy , dz ext{Integrate with respect to $x$.} = int_{z=0}^{z=1} int_{y=2}^{y=4} left[12+6yz^2 ight] ,dy , dz ext{Evaluate.} = int_{z=0}^{z=1} left[ left.12y+6dfrac{y^2}{2}z^2 ight|_{y=2}^{y=4} ight] dz ext{Integrate with respect to $y$.} = int_{z=0}^{z=1} [24+36z^2] , dz ext{Evaluate.} = left[ 24z+36dfrac{z^3}{3} ight]_{z=0}^{z=1} ext{Integrate with respect to $z$.} =36. ext{Evaluate.} end{align*}]

Example (PageIndex{2}): Evaluating a Triple Integral

Evaluate the triple integral

[iiint_B x^2 yz ,dV]

where (B = ig{(x,y,z),|, - 2 leq x leq 1, , 0 leq y leq 3, , 1 leq z leq 5 ig} ) as shown in Figure (PageIndex{2}).

Solution

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate (y) first, then (x), and then (z).

[egin{align*} iiintlimits_{B} x^2 yz ,dV = int_1^5 int_{-2}^1 int_0^3 [x^2 yz] ,dy , dx , dz = int_1^5 int_{-2}^1 left[ left. x^2 dfrac{y^3}{3} z ight|_0^3 ight] dx , dz = int_1^5 int_{-2}^1 dfrac{y}{2} x^2 z ,dx , dz = int_1^5 left[ left. dfrac{9}{2} dfrac{x^3}{3} z ight|_{-2}^1 ight] dz = int_1^5 dfrac{27}{2} z , dz = left. dfrac{27}{2} dfrac{z^2}{2} ight|_1^5 = 162. end{align*}]

Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to (x) first, then (z), then (y)

[egin{align*} iiiintlimits_{B} x^2yz ,dV = int_0^3 int_1^5 int_{-2}^1 [x^2yz] ,dx, dz, dy = int_0^3 int_1^5 left[ left. dfrac{x^3}{3} yz ight|_{-2}^1 ight] dz ,dy =int_0^3 int_1^5 3yz ; dz ,dy = int_0^3 left.left[ 3ydfrac{z^2}{2} ight|_1^5 ight] ,dy = int_0^3 36y ; dy = left. 36dfrac{y^2}{2} ight|_0^3 =18(9-0) =162. end{align*}]

Exercise (PageIndex{1})

Evaluate the triple integral

[iint_B z , sin , x , cos , y , dV onumber ]

where (B = ig{(x,y,z),|,0 leq x leq pi, , dfrac{3pi}{2} leq y leq 2pi, , 1 leq z leq 3 ig}).

Hint

Follow the steps in the previous example.

Answer

[iint_B z , sin , x , cos , y , dV = 8 onumber]

Triple Integral over a General Region

The triple integral of a continuous function (f(x,y,z)) over a general three-dimensional region

[E = ig{(x,y,z),|,(x,y) in D, , u_1(x,y) leq z leq u_2(x,y) ig}]

in (mathbb{R}^3), where (D) is the projection of (E) onto the (xy)-plane, is

[iiint_E f(x,y,z) ,dV = iint_D left[int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) ,dz ight] dA.]

Similarly, we can consider a general bounded region (D) in the (xy)-plane and two functions (y = u_1(x,z)) and (y = u_2(x,z)) such that (u_1(x,z) leq u_2(x,z)) for all (9x,z)) in (D). Then we can describe the solid region (E) in (mathbb{R}^3) as

[E = ig{(x,y,z),|,(x,z) in D, , u_1(x,z) leq z leq u_2(x,z) ig}] where (D) is the projection of (E) onto the (xy)-plane and the triple integral is

[iiint_E f(x,y,z),dV = iint_D left[int_{u_1(x,z)}^{u_2(x,z)} f(x,y,z) ,dy ight] dA.]

Finally, if (D) is a general bounded region in the (xy)-plane and we have two functions (x = u_1(y,z)) and (x = u_2(y,z)) such that (u_1(y,z) leq u_2(y,z)) for all ((y,z)) in (D), then the solid region (E) in (mathbb{R}^3) can be described as

[E = ig{(x,y,z),|,(y,z) in D, , u_1(y,z) leq z leq u_2(y,z) ig}] where (D) is the projection of (E) onto the (xy)-plane and the triple integral is

[iiint_E f(x,y,z),dV = iint_D left[int_{u_1(y,z)}^{u_2(y,z)} f(x,y,z) , dx ight] dA.]

Note that the region (D) in any of the planes may be of Type I or Type II as described in previously. If (D) in the (xy)-plane is of Type I (Figure (PageIndex{4})), then

[E = ig{(x,y,z),|,a leq x leq b, , g_1(x) leq y leq g_2(x), , u_1(x,y) leq z leq u_2(x,y) ig}.]

Then the triple integral becomes

[iiint_E f(x,y,z) ,dV = int_a^b int_{g_1(x)}^{g_2(x)} int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) ,dz , dy , dx.]

If (D) in the (xy)-plane is of Type II (Figure (PageIndex{5})), then

[E = ig{(x,y,z),|,c leq x leq d, h_1(x) leq y leq h_2(x), , u_1(x,y) leq z leq u_2(x,y) ig}.]

Then the triple integral becomes

[iiint_E f(x,y,z) ,dV = int_{y=c}^{y=d} int_{x=h_1(y)}^{x=h_2(y)} int_{z=u_1(x,y)}^{z=u_2(x,y)} f(x,y,z),dz , dx , dy.]

Example (PageIndex{3A}): Evaluating a Triple Integral over a General Bounded Region

Evaluate the triple integral of the function (f(x,y,z) = 5x - 3y) over the solid tetrahedron bounded by the planes (x = 0, , y = 0, , z = 0), and (x + y + z = 1).

Solution

Figure (PageIndex{6}) shows the solid tetrahedron (E) and its projection (D) on the (xy)-plane.

We can describe the solid region tetrahedron as

[E = ig{(x,y,z),|,0 leq x leq 1, , 0 leq y leq 1 - x, , 0 leq z leq 1 - x - y ig}. onumber]

Hence, the triple integral is

[iiint_E f(x,y,z) ,dV = int_{x=0}^{x=1} int_{y=0}^{y=1-x} int_{z=0}^{z=1-x-y} (5x - 3y) ,dz , dy , dx. onumber]

To simplify the calculation, first evaluate the integral (displaystyle int_{z=0}^{z=1-x-y} (5x - 3y) ,dz). We have

[int_{z=0}^{z=1-x-y} (5x - 3y) ,dz = (5x - 3y)z igg|_{z=0}^{z=1-x-y} = (5x - 3y)(1 - x - y). onumber]

Now evaluate the integral

[int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y) ,dy, onumber]

obtaining

[int_{y=0}^{y=1-x} (5x - 3y)(1 - x - y),dy = dfrac{1}{2}(x - 1)^2 (6x - 1). onumber]

Finally evaluate

[int_{x=0}^{x=1} dfrac{1}{2}(x - 1)^2 (6x - 1),dx = dfrac{1}{12}. onumber]

Putting it all together, we have

[iiint_E f(x,y,z),dV = int_{x=0}^{x=1} int_{y=0}^{y=1-x} int_{z=0}^{z=1-x-y}(5x - 3y),dz , dy , dx = dfrac{1}{12}. onumber]

Just as we used the double integral [iint_D 1 ,dA] to find the area of a general bounded region (D) we can use [iiint_E 1,dV] to find the volume of a general solid bounded region (E). The next example illustrates the method.

Example (PageIndex{3B}): Finding a Volume by Evaluating a Triple Integral

Find the volume of a right pyramid that has the square base in the (xy)-plane ([-1,1] imes [-1,1]) and vertex at the point ((0, 0, 1)) as shown in the following figure.

Solution

In this pyramid the value of (z) changes from 0 to 1 and at each height (z) the cross section of the pyramid for any value of (z) is the square

[[-1 + z, , 1 - z] imes [-1 + z, , 1 - z]. onumber]

Hence, the volume of the pyramid is [iiint_E 1,dV onumber] where

[E = ig{(x,y,z),|,0 leq z leq 1, , -1 + z leq y leq 1 - z, , -1 + z leq x leq 1 - z ig}. onumber]

Thus, we have

[egin{align*} iiint_E 1,dV = int_{z=0}^{z=1} int_{y=1+z}^{1-z} int_{x=1+z}^{1-z} 1,dx , dy , dz = int_{z=0}^{z=1} int_{y=1+z}^{1-z} (2 - 2z), dy , dz = int_{z=0}^{z=1}(2 - 2z)^2 ,dz = dfrac{4}{3}. end{align*}]

Hence, the volume of the pyramid is (dfrac{4}{3}) cubic units.

Exercise (PageIndex{3})

Consider the solid sphere (E = ig{(x,y,z),|,x^2 + y^2 + z^2 = 9 ig}). Write the triple integral [iiint_E f(x,y,z) ,dV onumber ] for an arbitrary function (f) as an iterated integral. Then evaluate this triple integral with (f(x,y,z) = 1). Notice that this gives the volume of a sphere using a triple integral.

Hint

Follow the steps in the previous example. Use symmetry.

Answer

[egin{align*} iiint_E 1,dV = 8 int_{x=-3}^{x=3} int_{y=-sqrt{9-z^2}}^{y=sqrt{9-z^2}}int_{z=-sqrt{9-x^2-y^2}}^{z=sqrt{9-x^2-y^2}} 1,dz , dy , dx = 36 pi , ext{cubic units}. end{align*}]

Changing the Order of Integration

As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.

Example (PageIndex{4}): Changing the Order of Integration

Consider the iterated integral

[int_{x=0}^{x=1} int_{y=0}^{y=x^2} int_{z=0}^{z=y} f(x,y,z),dz , dy , dx.]

The order of integration here is first with respect to z, then y, and then x. Express this integral by changing the order of integration to be first with respect to (x), then (z), and then (y). Verify that the value of the integral is the same if we let (f (x,y,z) =xyz).

Solution

The best way to do this is to sketch the region (E) and its projections onto each of the three coordinate planes. Thus, let

[E = ig{(x,y,z),|,0 leq x leq 1, , 0 leq y leq x^2, , 0 leq z leq y ig}. onumber]

and

[int_{x=0}^{x=1} int_{y=0}^{y=x^2} int_{z=0}^{z=x^2} f(x,y,z) ,dz , dy , dx = iiint_E f(x,y,z),dV. onumber]

We need to express this triple integral as

[int_{y=c}^{y=d} int_{z=v_1(y)}^{z=v_2(y)} int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z),dx , dz , dy. onumber]

Knowing the region (E) we can draw the following projections (Figure (PageIndex{8})):

on the (xy)-plane is (D_1 = ig{(x,y),|, 0 leq x leq 1, , 0 leq y leq x^2 ig} = { (x,y) ,|, 0 leq y leq 1, , sqrt{y} leq x leq 1 ig},)

on the (yz)-plane is (D_2 = ig{(y,z) ,|, 0 leq y leq 1, , 0 leq z leq y^2 ig}), and

on the (xz)-plane is (D_3 = ig{(x,z) ,|, 0 leq x leq 1, , 0 leq z leq x^2 ig}).

Now we can describe the same region (E) as (ig{(x,y,z) ,|, 0 leq y leq 1, , 0 leq z leq y^2, , sqrt{y} leq x leq 1 ig}), and consequently, the triple integral becomes

[int_{y=c}^{y=d} int_{z=v_1(y)}^{z=v_2(y)} int_{x=u_1(y,z)}^{x=u_2(y,z)} f(x,y,z),dx , dz , dy = int_{y=0}^{y=1} int_{z=0}^{z=x^2} int_{x=sqrt{y}}^{x=1} f(x,y,z),dx , dz , dy]

Now assume that (f (x,y,z) = xyz) in each of the integrals. Then we have

[egin{align*} int_{x=0}^{x=1} int_{y=0}^{y=x^2} int_{z=0}^{z=y^2} xyz , dz , dy , dx = int_{x=0}^{x=1} int_{y=0}^{y=x^2} left. left[xy dfrac{z^2}{2} ight|_{z=0}^{z=y^2} ight] , dy , dx = int_{x=0}^{x=1} int_{y=0}^{y=x^2} left( x dfrac{y^5}{2} ight) dy , dx = int_{x=0}^{x=1} left. left[ xdfrac{y^6}{12} ight|_{y=0}^{y=x^2} ight] dx = int_{x=0}^{x=1} dfrac{x^{13}}{12} dx = left. dfrac{x^{14}}{168} ight|_{x=0}^{x=1} = dfrac{1}{168}, end{align*}]

[ egin{align*} int_{y=0}^{y=1} int_{z=0}^{z=y^2} int_{x=sqrt{y}}^{x=1} xyz , dx , dz , dy = int_{y=0}^{y=1} int_{z=0}^{z=y^2} left.left[yz dfrac{x^2}{2} ight|_{sqrt{y}}^{1} ight] dz , dy = int_{y=0}^{y=1} int_{z=0}^{z=y^2} left( dfrac{yz}{2} - dfrac{y^2z}{2} ight) dz , dy = int_{y=0}^{y=1} left. left[ dfrac{yz^2}{4} - dfrac{y^2z^2}{4} ight|_{z=0}^{z=y^2} ight] dy = int_{y=0}^{y=1} left(dfrac{y^5}{4} - dfrac{y^6}{4} ight) dy = left. left(dfrac{y^6}{24} - dfrac{y^7}{28} ight) ight|_{y=0}^{y=1} = dfrac{1}{168}. end{align*} ]

The answers match.

Exercise (PageIndex{4})

Write five different iterated integrals equal to the given integral

[int_{z=0}^{z=4} int_{y=0}^{y=4-z} int_{x=0}^{x=sqrt{y}} f(x,y,z) , dx , dy , dz. onumber]

Hint

Follow the steps in the previous example, using the region (E) as ( ig{(x,y,z) ,|, 0 leq z leq 4, , 0 leq y leq 4 - z, , 0 leq x leq sqrt{y} ig}), and describe and sketch the projections onto each of the three planes, five different times.

Answer

[(i) , int_{z=0}^{z=4} int_{x=0}^{x=sqrt{4-z}} int_{y=x^2}^{y=4-z} f(x,y,z) , dy , dx , dz, , (ii) , int_{y=0}^{y=4} int_{z=0}^{z=4-y} int_{x=0}^{x=sqrt{y}} f(x,y,z) ,dx , dz , dy, ,(iii) , int_{y=0}^{y=4} int_{x=0}^{x=sqrt{y}} int_{z=0}^{Z=4-y} f(x,y,z) ,dz , dx , dy, , onumber]

[ (iv) , int_{x=0}^{x=2} int_{y=x^2}^{y=4} int_{z=0}^{z=4-y} f(x,y,z) ,dz , dy , dx, , (v) int_{x=0}^{x=2} int_{z=0}^{z=4-x^2} int_{y=x^2}^{y=4-z} f(x,y,z) ,dy , dz , dx onumber]

Example (PageIndex{5}): Changing Integration Order and Coordinate Systems

Evaluate the triple integral

[iiint_{E} sqrt{x^2 + z^2} ,dV, onumber ]

where (E) is the region bounded by the paraboloid (y = x^2 + z^2) (Figure (PageIndex{9})) and the plane (y = 4).

Solution

The projection of the solid region (E) onto the (xy)-plane is the region bounded above by (y = 4) and below by the parabola (y = x^2) as shown.

Thus, we have

[E = ig{(x,y,z) ,|, -2 leq x leq 2, , x^2 leq y leq 4, , -sqrt{y - x^2} leq z sqrt{y - x^2} ig}. onumber]

The triple integral becomes

[iiint_E sqrt{x^2 + z^2} ,dV = int_{x=-2}^{x=2} int_{y=x^2}^{y=4} int_{z=-sqrt{y-x^2}}^{z=sqrt{y-x^2}} sqrt{x^2 + z^2} ,dz , dy , dx. onumber]

This expression is difficult to compute, so consider the projection of (E) onto the (xz)-plane. This is a circular disc (x^2 + z^2 leq 4). So we obtain

[iiint_E sqrt{x^2 + z^2} ,dV = int_{x=-2}^{x=2} int_{y=x^2}^{y=4} int_{z=-sqrt{y-x^2}}^{z=sqrt{y-x^2}} sqrt{x^2 + z^2} ,dz , dy , dx = int_{x=-2}^{x=2} int_{z=-sqrt{4-x^2}}^{z=sqrt{4-x^2}} int_{y=x^2+z^2}^{y=4} sqrt{x^2 + z^2} ,dy , dz , dx. onumber]

Here the order of integration changes from being first with respect to (z) then (y) and then (x) to being first with respect to (y) then to (z) and then to (x). It will soon be clear how this change can be beneficial for computation. We have

[int_{x=-2}^{x=2} int_{z=sqrt{4-x^2}}^{z=sqrt{4-x^2}} int_{y=x^2+z^2}^{y=4} sqrt{x^2 + z^2} ,dy , dz , dx = int_{x=-2}^{x=2} int_{z=-sqrt{4-x^2}}^{z=sqrt{4-x^2}} (4 - x^2 - z^2) sqrt{x^2 + z^2} ,dz , dx. onumber]

Now use the polar substitution (x = r , cos , heta, , z = r , sin , heta), and (dz , dx = r , dr , d heta) in the (xz)-plane. This is essentially the same thing as when we used polar coordinates in the (xy)-plane, except we are replacing (y) by (z). Consequently the limits of integration change and we have, by using (r^2 = x^2 + z^2),

[int_{x=-2}^{x=2} int_{z=-sqrt{4-x^2}}^{z=sqrt{4-x^2}} (4 - x^2 - z^2) sqrt{x^2 + z^2},dz , dx = int_{ heta=0}^{ heta=2pi} int_{r=0}^{r=2} (4 - r^2) rr , dr , d heta = int_0^{2pi} left. left[ dfrac{4r^3}{3} - dfrac{r^5}{5} ight|_0^2 ight] , d heta = int_0^{2pi} dfrac{64}{15} ,d heta = dfrac{128pi}{15} onumber]

Average Value of a Function of Three Variables

Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.

Average Value of a Function of Three Variables

If (f(x,y,z)) is integrable over a solid bounded region (E) with positive volume (V , (E),) then the average value of the function is

[f_{ave} = dfrac{1}{V , (E)} iiint_E f(x,y,z) , dV.]

Note that the volume is

[V , (E) = iiint_E 1 ,dV.]

Example (PageIndex{6}): Finding an Average Temperature

The temperature at a point ((x,y,z)) of a solid (E) bounded by the coordinate planes and the plane (x + y + z = 1) is (T(x,y,z) = (xy + 8z + 20) , ext{°} ext{C} ). Find the average temperature over the solid.

Solution

Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane (x + y + z = 1) has intercepts ((1,0,0), , (0,1,0),) and ((0,0,1)). The region (E) looks like

[E = ig{(x,y,z) ,|, 0 leq x leq 1, , 0 leq y leq 1 - x, , 0 leq z leq 1 - x - y ig}. onumber]

Hence the triple integral of the temperature is

[iiint_E f(x,y,z) ,dV = int_{x=0}^{x=1} int_{y=0}^{y=1-x} int_{z=0}^{z=1-x-y} (xy + 8z + 20) , dz , dy , dx = dfrac{147}{40}. onumber ]

The volume evaluation is

[V , (E) = iiint_E 1,dV = int_{x=0}^{x=1} int_{y=0}^{y=1-x} int_{z=0}^{z=1-x-y} 1 ,dz , dy , dx = dfrac{1}{6}. onumber ]

Hence the average value is

[ T_{ave} = dfrac{147/40}{1/6} = dfrac{6(147)}{40} = dfrac{441}{20} , ext{°} ext{C} onumber].

Exercise (PageIndex{6})

Find the average value of the function (f(x,y,z) = xyz) over the cube with sides of length 4 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.

Hint

Follow the steps in the previous example.

Answer

(f_{ave} = 8)

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


Evaluate the triple integral.

Evaluate the triple integral of $x=y^2$ over the region bounded by $z=x$, $z=0$ and $x=1$ My order of integration was $dx:dy:dz$.

I want to calculate the volume of this surface. I solved it for $dz:dy:dx$ and it was:

And for $dz:dx:dy$ would be this:

I tried to solve it and the result is this:

But i think its wrong please advice me the best solution .

I wanted to post the shape of this surface in 3-dimensional region but I couldn't because I am new user.


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Contents

By polar coordinates Edit

A standard way to compute the Gaussian integral, the idea of which goes back to Poisson, [3] is to make use of the property that:

Comparing these two computations yields the integral, though one should take care about the improper integrals involved.

where the factor of r is the Jacobian determinant which appears because of the transform to polar coordinates (r dr is the standard measure on the plane, expressed in polar coordinates Wikibooks:Calculus/Polar Integration#Generalization), and the substitution involves taking s = −r 2 , so ds = −2r dr.

Complete proof Edit

To justify the improper double integrals and equating the two expressions, we begin with an approximating function:

were absolutely convergent we would have that its Cauchy principal value, that is, the limit

To see that this is the case, consider that

Using Fubini's theorem, the above double integral can be seen as an area integral

Since the exponential function is greater than 0 for all real numbers, it then follows that the integral taken over the square's incircle must be less than I ( a ) 2 > , and similarly the integral taken over the square's circumcircle must be greater than I ( a ) 2 > . The integrals over the two disks can easily be computed by switching from cartesian coordinates to polar coordinates:

By the squeeze theorem, this gives the Gaussian integral

By Cartesian coordinates Edit

A different technique, which goes back to Laplace (1812), [3] is the following. Let

Since the limits on s as y → ±∞ depend on the sign of x, it simplifies the calculation to use the fact that ex 2 is an even function, and, therefore, the integral over all real numbers is just twice the integral from zero to infinity. That is,

Thus, over the range of integration, x ≥ 0, and the variables y and s have the same limits. This yields:

The integral of a Gaussian function Edit

The integral of an arbitrary Gaussian function is

This form is useful for calculating expectations of some continuous probability distributions related to the normal distribution, such as the log-normal distribution, for example.

N-dimensional and functional generalization Edit

Suppose A is a symmetric positive-definite (hence invertible) n × n precision matrix, which is the matrix inverse of the covariance matrix. Then,

where the integral is understood to be over R n . This fact is applied in the study of the multivariate normal distribution.

where σ is a permutation of <1, . 2N> and the extra factor on the right-hand side is the sum over all combinatorial pairings of <1, . 2N> of N copies of A −1 .

for some analytic function f, provided it satisfies some appropriate bounds on its growth and some other technical criteria. (It works for some functions and fails for others. Polynomials are fine.) The exponential over a differential operator is understood as a power series.

While functional integrals have no rigorous definition (or even a nonrigorous computational one in most cases), we can define a Gaussian functional integral in analogy to the finite-dimensional case. [ citation needed ] There is still the problem, though, that ( 2 π ) ∞ > is infinite and also, the functional determinant would also be infinite in general. This can be taken care of if we only consider ratios:

In the DeWitt notation, the equation looks identical to the finite-dimensional case.

N-dimensional with linear term Edit

If A is again a symmetric positive-definite matrix, then (assuming all are column vectors)

Integrals of similar form Edit

One could also integrate by parts and find a recurrence relation to solve this.

Higher-order polynomials Edit

Applying a linear change of basis shows that the integral of the exponential of a homogeneous polynomial in n variables may depend only on SL(n)-invariants of the polynomial. One such invariant is the discriminant, zeros of which mark the singularities of the integral. However, the integral may also depend on other invariants. [5]

Exponentials of other even polynomials can numerically be solved using series. These may be interpreted as formal calculations when there is no convergence. For example, the solution to the integral of the exponential of a quartic polynomial is [ citation needed ]

The n + p = 0 mod 2 requirement is because the integral from −∞ to 0 contributes a factor of (−1) n+p /2 to each term, while the integral from 0 to +∞ contributes a factor of 1/2 to each term. These integrals turn up in subjects such as quantum field theory.


Introduction

Triple integrals are more difficult (in fact, usually impossible) to picture than double integrals, but the principle is much the same. Instead of integrals like ∬ R f ( x , y ) d A , iint_R f(x,y) dA, ∬ R ​ f ( x , y ) d A , now we'll handle integrals like ∭ D f ( x , y , z ) d V iiint_D f(x,y,z) dV ∭ D ​ f ( x , y , z ) d V where D D D is a 3 3 3 -dimensional body.

Again, the challenge with triple integrals really lies with finding the limits of integration once we have those, we simply have to carry out the integration, which usually isn't too complicated.

Evaluate the following triple integral. ∫ 0 2 ∫ 0 3 ∫ 0 2 d x d y d x int_0^2 int_0^3 int_0^2 dx dy dx ∫ 0 2 ​ ∫ 0 3 ​ ∫ 0 2 ​ d x d y d x

Solution

Note that this gives the volume of a 2 × 3 × 2 2×3×2 2 × 3 × 2 cube.

Solution

Notice that in each case, we have to make sure that we use the limits that involve one variable before we integrate with respect to that variable.

Find the volume of the solid in the first octant bounded by the plane 2 x + 3 y + 6 z = 12 2x+3y+6z=12 2 x + 3 y + 6 z = 1 2 and the coordinate planes.


Single Integrals

The Quad function is the workhorse of SciPy’s integration functions. Numerical integration is sometimes called quadrature, hence the name. It is normally the default choice for performing single integrals of a function f(x) over a given fixed range from a to b.

The general form of quad is scipy.integrate.quad(f, a, b), Where ‘f’ is the name of the function to be integrated. Whereas, ‘a’ and ‘b’ are the lower and upper limits, respectively. Let us see an example of the Gaussian function, integrated over a range of 0 and 1.

We first need to define the function &rarr $f(x) = e^<-x^2>$ , this can be done using a lambda expression and then call the quad method on that function.

The above program will generate the following output.

The quad function returns the two values, in which the first number is the value of integral and the second value is the estimate of the absolute error in the value of integral.

Note &minus Since quad requires the function as the first argument, we cannot directly pass exp as the argument. The Quad function accepts positive and negative infinity as limits. The Quad function can integrate standard predefined NumPy functions of a single variable, such as exp, sin and cos.


Calculating the Bounds

In the next part, which differs from the steps above for the indefinite integral, you’re looking for a specific area. For this example, it’s the area between 0 and 4 on the graph (i.e. the bounds of integration specified in the problem):

Step 1: Calculate the integral at the upper bound (which is 4):

Step 2: Calculate the integral at the lower bound (which is 0):

Step 3: Calculate the final answer by subtracting the parts (upper bound – lower bound):


4.5: Triple Integrals

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tripleintegral.zip
Filename tripleintegral.zip ( Download )
Title Triple Integral
Description This program will compute triple integrals using composite Simpson rule. It can run slow, but it is often accurate. Please read documentation. Enjoy!
Author Jeremy Lane ( [email protected] )
Category TI-83/84 Plus BASIC Math Programs (Calculus)
File Size 1,195 bytes
File Date and Time Fri Jul 22 21:45:13 2016
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Calculus is known as one of the critical fields of study in Mathematics. It is the study of continuous change. The branch of Calculus emphasizes the concepts of Limits, Functions, Integrals, Infinite series, and Derivatives. Limits is one of the essential concepts of calculus. It helps in analyzing the value of a function or sequence approaches as the input or index approaches a particular point. In other words, it depicts how any function acts near a point and not at that given point. The theory of Limits lays a foundation for Calculus it used to define Continuity, Integrals, and Derivatives.

Limits are stated for a function, any discrete sequence, and even real-valued function or complex functions. For a function f(x) , the value the function takes as the variable approaches a specific number say n then x → n is known as the limit. Here the function has a finite limit:

Where, L= Lim x → x0 f(x) for point x0 . For all ε > 0 we can find δ > 0 where absolute value of f(x) – L is less than E when absolute value of x - x0 . In the case of a sequence of real numbers, like a1, a2, a3,…, an. The real number L is the limit of the sequence:

The value of the function f(x) can be found from the left or the right of the point n . The expected value of the function for the points to the left of the given point n is the left-hand limit, also called the below limit, while the points to the right of the specified point n is known as the right-hand limit even called the above limit. The limit on the left is defined by limx → x- 0 f(x) and the limit on the right is denoted by limx → x + 0 f(x) .

It is important to understand that the limit exists only when the values derived for the left-hand limit and the right-hand limit are equal. While calculating the limit for complex-figured functions, there are unlimited modes to approach a limit for a point. In such situations to find a distinct value of the limit, there is a need forstricter standards. For the limit of a rational function of the type p(x) / q(x) , the important step is to simplify the rational function to the form 0/0 for a given point.

There are various ways for the computation of limits depending on the different nature and types of functions. There is a great application of the L- Hospital's rule, which involves differentiating the numerator and denominator of rational functions or indeterminable limits, till the limit takes the form 0/0 or ∞/∞ .


Change of Variables in Multiple Integrals

Substitution (or change of variables) is a powerful technique for evaluating integrals in single variable calculus. An equivalent transformation is available for dealing with multiple integrals. The idea is to replace the original variables of integration by the new set of variables. This way the integrand is changed as well as the bounds for integration. If we are lucky enough to find a convenient change of variables we can significantly simplify the integrand or the bounds.

Change of variables formula

Two dimensional pictures are the easiest to draw so we will start with functions of two variables. Our first task is to get familiar with transformations of two dimensional regions.

Transformations in ( mathbb R^2 )

Assume that ( S ) is a region in ( mathbb R^2 ). We want to study the ways in which this region can be transformed to another region ( T ).

This is easiest to explain by considering an example. Let ( S=[0,2] imes[0,2] ). Consider the functions ( u:S o mathbb R ) and ( v:S omathbb R ) defined in the following way: egin u(x,y)&=&x+2y v(x,y)&=&x-y.end To every point ( (x,y)in S ) (( S ) is painted blue in the diagram on the left) we can assign a new green point with coordinates ( (u(x,y),v(x,y)) ). This way we obtain a green region ( T ).

The mapping ( (x,y)mapsto (u(x,y),v(x,y)) ) is one-to-one and onto, hence a bijection (you may want to review these terms in the section Functions). We can also write the inverse transformation, that maps each point ( (u,v)in T ) to the point ( (x(u,v),y(u,v)) ) in the following way: egin x(u,v)&=&frac3 y(u,v)&=&frac3. end

Change of variables in double integrals

Assume that ( Ssubseteq mathbb R^2 ) is a region in the plane. Let ( Tsubseteqmathbb R^2 ) be another region and assume that there are continuously differentiable functions ( X:T omathbb R ) and ( Y:T omathbb R ), such that the mapping ( Phi(u,v)= (X(u,v),Y(u,v)) ) is a bijection between ( T ) and ( S ).

We omit the proof - you’ll do quite fine in calculus without knowing this proof.

The transformation ( (x,y)mapsto (2x+3y,x-3y) ) is linear, and the image of the original parallelogram must be parallelogram as well. The vertex ( (0,0) ) gets mapped to the vertex ( (0,0) ) of the new parallelogram. Similarly, ( left(1,frac13 ight) ) gets mapped to ( (3,0) ), ( left(frac43,frac19 ight)mapsto (3,1) ), and ( left(frac13,-frac29 ight)mapsto (0,1) ).

In order to find the Jacobian we need to express ( x ) and ( y ) in terms of ( u ) and ( v ). We do this by solving the system ( u=x+3y ), ( v=y ), and we get:

Denote by ( E ) the parallelogram with vertices ( (0,0) ), ( (3,0) ), ( (3,1) ), and ( (0,1) ). We have eginiint_D e^<2x+3y>cdot cos(x-3y),dxdy&=&iint_E e^ucdot cos v cdotleft|-frac19 ight|,dudv=frac19iint_Ee^ucos v,dudv &=&frac19int_0^1int_0^3e^ucos v,dudv=frac19 int_0^1left.left(cos vcdot e^u ight) ight|_^,dv &=&frac19int_0^1 left(e^3-1 ight)cdot cos v,dv=frac9cdot left.sin v ight|_^ &=&frac<9>.end

Change of variables in triple integrals

Assume that ( S, Tsubseteq mathbb R^3 ) are two regions in space. Assume that there are continuously differentiable functions ( X:T omathbb R ), ( Y:T omathbb R ), and ( Z:T omathbb R ), such that the mapping ( Phi:T o S ) defined as ( Phi(u,v,w)= (X(u,v,w),Y(u,v,w),Z(u,v,w)) ) is a bijection.

Omitted. See any real analysis textbook.

Polar, cylindrical, and spherical substitutions

We will now study very important substitutions that are used to simplify integrations over circular, spherical, cylindrical, and elliptical domains. One of them is applicable to double integral and is called polar change of variables and the other two, cylindrical and spherical , are used in triple integralds.

Polar substitution

The following change of variables is called the polar substitution: egin x&=&rcos heta y&=&rsin heta. end The Jacobian for the polar substitution is equal to: [ frac=detleft|egin cos heta&-rsin heta sin heta&rcos hetaend ight|=rcos^2 heta+rsin^2 heta=r.]

The variables ( r ) and ( heta ) have the geometric meaning in the ( xy )-coordinate system. The distance between ( (x,y) ) and the origin is precisely ( r ), while ( heta ) is the angles between the ( x )-axis and the line connecting ( (x,y) ) with ( (0,0) ).

Let us use the following substitution: egin x&=&rcos heta y&=&rsin heta 0leq &r&leq 3 0leq & heta &< 2pi. end The transformation ( (r, heta)mapsto (rcos heta,rsin heta) ) is a bijection between the rectangle ( [0,3] imes[0,2pi] ) in the ( (r, heta) )-plane and the disc of radius ( 3 ).

Since ( x^2+y^2=r^2cos^2 heta+r^2sin^2 heta=r^2 ), the integral becomes: egin iint_D cosleft(x^2+y^2 ight),dxdy&=&int_0^<2pi>int_0^3 cosleft(r^2 ight) cdot r,drd heta. end For the integral ( int_0^3cosleft(r^2 ight)r,dr ) we use the substitution ( r^2=u ). Then we have ( r=sqrt u ) and ( dr=frac1<2sqrt u>,du ). The bounds of integration become ( 0leq uleq 9 ), and the integral is [ int_0^3cos(r^2)r,dr=int_0^9cos ucdot sqrt ucdot frac1<2sqrt u>,du=frac12int_0^9cos u,du=frac2.] Therefore, [ iint_D cosleft(x^2+y^2 ight),dxdy= int_0^<2pi>frac2,d heta= sin 9cdot pi.]

When dealing with ellipses it is very common to use the modified polar substitution. If the equation of the ellipse is ( frac+frac=1 ), the following substitution is used to describe its interior: egin x&=&arcos heta y&=&brsin heta 0leq&r&leq 1 0leq& heta&leq 2pi. end

Let us use the following substitution: egin x&=&rcos heta y&=&rsin heta z&=&z 0leq &r&leq 2 0leq & heta &< frac4 0leq&z&leq 4-r^2. end The transformation ( (r, heta,z)mapsto (rcos heta,rsin heta,z) ) is a bijection between the solid the solid ( B ) defined as: [ B=left<(r, heta,z):0leq rleq 2, 0leq hetaleq frac4, 0leq z leq 4-r^2 ight>] in the ( (r, heta,z) )-space and the solid ( D ) from the formulation of the problem.

Since ( x^2+y^2=r^2cos^2 heta+r^2sin^2 heta=r^2 ), the integral becomes: egin iiint_D e^,dxdydz&=&int_0^<2>int_0^4>int_0^ <4-r^2>e^ cdot r,dzd heta dr &=&int_0^<2>int_0^4> e^ cdot r(4-r^2),d heta dr &=&frac4cdotint_0^2 r(4-r^2)e^,dr. end In the last integral we use the substitution ( r^2=u ). Then we have ( r=sqrt u ) and ( dr=frac1<2sqrt u>,du ). The bounds of integration become ( 0leq uleq 4 ), and the integral is [ int_0^2 r(4-r^2)e^,dr=int_0^4sqrt ucdot(4-u)cdot e^ucdotfrac1<2sqrt u>,du=frac12int_0^44e^u,du-frac12int_0^4ue^u,du.] The first term on the right-hand side is equal to ( 2left(e^4-1 ight) ), and for the second we use the integration by parts. We take ( f=u ), ( dg=e^udu ), which gives us ( g=e^u ) and the integral becomes: [ int_0^4ue^u,du=left.ue^u ight|_0^4-int_0^4e^u,du=4e^4-e^4+1.] Therefore [ int_0^2r(4-r^2)e^,dr=2e^4-2-2e^4+frac2-frac12=frac2,] thus [ iiint_D e^,dxdydz= frac8. ]

Spherical substitution

Spherical substitution means replacing the original variables ( (x,y,z) ) by the variables ( ( ho, heta, phi) ), where ( ho ) is the distance of the points ( (x,y,z) ) from the origin ( (0,0,0) ) ( heta ) is the angle that the line connecting ( (0,0,0) ) and ( (x,y,0) ) forms with the ( x )-axis, and ( phi ) is the angle between the ( z )-axis and the line connecting ( (x,y,z) ) with ( (0,0,0) ). Mathematically, the equations are: egin x&=& hocos hetasinphi y&=& hosin hetasinphi z&=& hocosphi. end We can find the Jacobian by calculating the appropriate determinant: egin frac&=&detleft|egin cos hetasinphi &- hosin hetasinphi & hocos hetacosphi sin hetasinphi& hocos hetasinphi& hosin hetacosphi cosphi&0&- hosinphiend ight| &=&- ho^2cos^2 hetasin^3phi- ho^2sin^2 hetasinphicos^2phi- ho^2cos^2 hetasinphicos^2phi- ho^2sin^2 hetasin^3phi &=&- ho^2sin^3phi- ho^2sinphicos^2phi=- ho^2sinphi .end Since in evaluation of the integral we are using the absolute value of the Jacobian, and ( phiinleft(0,frac2 ight) ) it is sufficient and more convenient to remember that [ left|frac ight|= ho^2sinphi.]

Let us use the following substitution: egin x&=& hocos hetasinphi y&=& hosin hetasinphi z&=& hocosphi 0leq & ho&leq 3 0leq & heta &< frac4 0leq&phi&leq frac2. end The evaluation of the integral is now easy as it becomes an iterated integral in variables ( ho ), ( heta ), and ( phi ). egin iiint_D e^>,dV&=& int_0^<3>int_0^4>int_0^2> e^< ho>cdot ho^2cdotsinphi,dphi d heta d ho =int_0^3int_0^4>e^< ho>cdot ho^2cdot left.left(-cosphi ight) ight|_^2>,d heta d ho &=&int_0^3int_0^4>e^< ho>cdot ho^2cdot 1,d heta d ho=frac4cdotint_0^3 ho^2e^< ho>,d ho.end In the last integral we can use integration by parts with functions ( u= ho^2 ), ( d v=e^< ho>,d ho ). Then we have ( du=2 ho,d ho ) and we can take ( v=e^ < ho>). The integral becomes: egin int_0^3e^< ho> ho^2,d ho=left. ho^2e^< ho> ight|_0^3-2int_0^3 ho e^< ho>,d ho &=& 9e^3-left.2 ho e^< ho> ight|0^<3>+2int_0^3e^< ho>,d ho=9e^3-6e^3+2e^3-2=5e^3-2. end Thus the final result is: egin iiint_D e^>,dV&=&frac4. end


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