# Set-up a triple integral in spherical coordinates of a solid bounded by a hemisphere and cylinder

Given $$S$$ as the solid that is outside the cylinder $$x^{2}+y^{2}=1,$$ inside the sphere $$x^{2}+y^{2}+z^{2}=4,$$ and above the $$x y$$ -plane. Set-up an iterated triple integral in spherical coordinates that is equal to $$\iiint_{G} f(x, y, z) d V$$ where $$f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$$.

How I approached this problem is that since

$$f(x, y, z)=x \sqrt{x^{2}+y^{2}+z^{2}}$$ and we know that from rectangular to spherical we have $$\rho=\sqrt{x^2+y^2+z^2}$$ and $$x=\rho sin\phi cos\theta$$

I know have $$\iiint_{G} f(x, y, z) d V = \iiint_{G} \rho sin\phi cos\theta*\rho^2 sin\phi d\rho d\theta d\phi$$

I've also determined the bounds for $$\theta = 0$$,$$2\pi$$ as well as for $$\phi = 0$$,$$\pi/2$$

My problem is determining the bounds for $$\rho$$ as the inner figure is a cylinder. My guess is that the upper bound of $$\rho=2$$ but I can't determine the lower bound.

Any help would be appreciated. Thanks.

• Do you have to set up the integral or just solve it? The integrand is an odd function and the region has $x\leftrightarrow -x$ symmetry so the integral is $0$ Dec 16 '20 at 17:20
• Just set-up the integral no need to evaluate it. I tried doing cylindrical coordinates to spherical but I'm having a hard time processing it and I think it isn't the right way to look at it Dec 16 '20 at 17:21

You are right about the outer bound, the inner bound will have to be the cylinder:

$$x^2+y^2=1 \implies \rho\sin\phi = 1$$

In the $$\rho\phi$$-plane ($$\rho$$ being the vertical axis) the region of integration will be shaded area to the left of the green line:

You can see from the photo that $$\phi$$ doesn't actually make it all the way back to $$0$$ (the $$z$$ axis in Cartesian), it stops at

$$2\sin\phi = 1 \implies \phi = \frac{\pi}{6}$$

and $$\rho$$ stops at

$$\rho\sin \frac{\pi}{2} = 1 \implies \rho = 1$$

From here it's easier to see how to set up the integral in both orders.

$$\int_0^{2\pi}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int_{\csc\phi}^{2} f\cdot \rho^2\sin\phi \: d\rho d\phi d\theta$$

$$\int_0^{2\pi}\int_{1}^{2}\int_{\sin^{-1}\left(\frac{1}{\rho}\right)}^{\frac{\pi}{2}} f\cdot \rho^2\sin\phi \: d\phi d\rho d\theta$$

• @Damian no that is not correct. I already provided the correct bounds in my answer and explained why those bounds were wrong. Dec 16 '20 at 18:23

The intersection of cylinder and sphere is at $$z = \sqrt 3$$

So $$\phi$$ at that point is $$\phi = \arccos (\frac{z}{2}) = \frac{\pi}{6}$$. (radius of sphere is $$2$$)

So $$\frac{\pi}{6} \leq \phi \leq \frac{\pi}{2}$$

Also $$\frac{1}{\sin \phi} \leq \rho \leq 2$$

Take a radial line from the origin to the point on the sphere and as the radius of the cylinder is $$1$$ the hypotenuse from origin to the point on the cylinder will be $$OT = \frac{OI}{\cos(90^0-\phi)} = \frac{1}{\sin \phi}$$