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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.

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  • $\begingroup$ 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 $0$ $\endgroup$ Dec 16 '20 at 17:20
  • $\begingroup$ 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 $\endgroup$
    – Damian
    Dec 16 '20 at 17:21
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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: enter image description here

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.

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

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

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  • $\begingroup$ @Damian no that is not correct. I already provided the correct bounds in my answer and explained why those bounds were wrong. $\endgroup$ Dec 16 '20 at 18:23
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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 $OT = \frac{OI}{\cos(90^0-\phi)} = \frac{1}{\sin \phi}$

enter image description here

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