Math 225: Calculus III    Quiz 8    4/1&3/97

<<Calculus`Math225`

Set up a triple integral in cylindrical coordinates that gives the integral of the function f[x_,y_,z_] = x y over the solid region between the paraboloid 2x^2 + y^2 + z == 8 and the parabolic sheet z = y^2. (Do not evaluate the integral).

Solution

The top surface is z == 8 - 2x^2 - y^2 and the bottom surface is z == y^2. The common region over which these surfaces are graphs is given by the intersection equation,
    8 - 2x^2 - y^2 == z == y^2
or x^2 + y^2 == 4,  a circle of radius 2. Switching to cylindrical coordinates, we must replace x by r Cos[t] and y by r Sin[t] and add an extra factor of r to the integral. The disk of radius 2 is described by 0 <= r <= 2, 0 <= t <= 2Pi.

Integrate[ r Cos[t] r Sin[t] r, {t, 0, 2Pi}, {r, 0, 2},
{z, r^2Sin[t]^2, 8-2r^2Cos[t]^2-r^2Sin[t]^2}]

This value 0 is easy to predict, since by symmetry, for every point, say {a,b}, in the solid that gives a positive value of the funtion f[a,b] == a b there is another point in the region, {-a,b},  that gives the  opposite negative value, f[-a,b] == -a b. Although you did not have to do the integral, it was straightforward to carry out:

Integrate[ r Cos[t] r Sin[t] r,
{z, r^2Sin[t]^2, 8-2r^2Cos[t]^2-r^2Sin[t]^2}]

%//Simplify

Integrate[%,{r, 0, 2}]

Set up a triple integral in spherical coordinates that gives the integral of the function f[x_,y_,z_] = x y over the solid region inside the  sphere x^2 + y^2 + z^2 == 9 below the xy-plane. (Do not evaluate the integral).

Solution

sph = {x->r Cos[t]Sin[p], y-> r Sin[t]Sin[p], z-> r Cos[p]};

ParametricPlot3D[{x,y,z}/.sph/.r->3, {p,Pi/2,Pi},{t,0,2Pi}]

The region is defined by 0 <= r <= 3, Pi/2 <= p <= Pi, and 0 <= t <= 2Pi. We must add a factor of r^2 Sin[p] to the integral for the volume element.

Integrate[(x y /. sph)* r^2 Sin[p], {t,0,2Pi}, {p,Pi/2,Pi}, {r,0,3}]

Step by step:

(x y /. sph)* r^2 Sin[p]

Integrate[(x y /. sph)* r^2 Sin[p], {r,0,3}]

Integrate[%, {p,Pi/2,Pi}]

Integrate[%, {t,0,2Pi}]