triple integral, spherical co-ordinates

**Dr Zoidburg** · May 4th 2010, 12:04 AM

This question concerns the integral $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}z dzdxdy$
(a) Sketch or describe in words the domain of integration.
(b) Rewrite the integral in cylindrical coordinates.
(c) Rewrite the integral in spherical coordinates.
(d) Which version of the integral would you choose as being the easiest to evaluate?
Evaluate the integral using the version you have chosen.
Is this correct for (b)?
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r^2}^{\s qrt{8-r^2}}z dzdrd\theta$
hope it is!
Can anyone help me with (a) and (c)? If I can get (c), I should be able to do (d).

This is as far as I've got with part (a):
looking at the limits for x and y:
$0 \leq y \leq 2$ and $0\leq x \leq \sqrt{4-y^2}$
which means upper limit of x is $x = \sqrt{4-y^2}$
solving this: $x^2 + y^2 = 4$ , which is a circle with radius 2.
Thus D is a 1/4 circle radius 2 centered around the origin. But how do I get to the next part and incorporate z?

For part (c), since it's a 1/4 circle am I right in thinking that $0 \leq \theta \leq \frac{\pi}{2}$ and $0 \leq \phi \leq \frac{\pi}{2}$ ? This is just vague intuition, so if true how do I show this mathematically?

As always, thanks for any and all help

**HallsofIvy** · May 4th 2010, 02:40 AM

Start by drawing a picture. (Or as well as you can in 3d- I never could draw well!)

The integral you have is $\int_{0}^{2}\int_{0}^{\sqrt{4-y^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{8-x^2-y^2}}z dzdxdy$

The outer integral has y going from y= 0 to y= 2 so draw two horizontal lines there. The second integral has x going from x= 0 to $x= \sqrt{4- y^2}$ . The first is the y-axis and the second, since $x^2= 4- y^2$ is the same as $x^2+ y^2= 4$ , is the semi-circle with center at (0, 0), radius 2, and x> 0. This much tells us that $\theta$ goes from $-\pi/2$ to $\pi/2$ . Finally, $z= \sqrt{x^2+ y^2}$ is the same as $z^2= x^2+ y^2$ , a cone with $\rho^2 cos^2(\phi)= \rho^2 cos^2(\theta)sij^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)= \rho^2 sin^2(\phi)$ which reduces to [tex] which reduces to $\phi= \pm \pi/4$ , and $z= \sqrt{8- x^2- y^2}$ gives $x^2+ y^2+ z^2= 8$ , a hemi-sphere with center at (0, 0, 0), radius $\sqrt{8}= 2\sqrt{2}$ , and $z\ge 0$ , $\rho^2= 8$ so $\rho= \pm 2\sqrt{2}$ .

**Dr Zoidburg** · May 4th 2010, 06:48 PM

thanks for all that. A couple of things:
Shouldn't $0 \leq \theta \leq \frac{\pi}{2}$ since x and y are bounded by 0?
and why/how did you get $\phi = \pm \frac{\pi}{4}$ ? I thought it would be pi/2. (again, would it be between 0 and pi/4?)

**Dr Zoidburg** · May 4th 2010, 11:02 PM

I've been working on this problem this morning and think I've gotten closer to finishing the blasted thing.
Is it accurate to describe the shape as a sort of 'ice-cream cone'? That is a cone cut from a sphere centered at the origin.

And are these co-ordinates correct:
(b)cylindrical co-ordinates
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{r}^{\sqr t{8-r^2}}z dzdrd\theta$

(c)spherical co-ordinates
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{4}}\i nt_{}^{2\sqrt{2}}\rho.cos (\phi) d\rho. d\phi. d\theta$
But I can't get the lower limit for $\rho$ . I know $r=\rho. sin(\phi)$ so is this the lower limit?

finally, is the answer $\frac{8\pi}{3}$

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