Converting to spherical coordinates

Question

I have the triple integral: $${\int_{-1}^1} {\int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}} {\int_{- \sqrt{x^2 + y^2}}^1} dzdydx$$

I want to evaluate this integral by changing to spherical coordinates.

I'm relatively new to spherical coordinates but I have a basic idea of the overall process I'm supposed to use.

I need to find values for $\rho$, $\theta$, and $r$ using the bounds of the integral I currently have.

I tried setting $z = -\sqrt{x^2 +y^2}$ which then means $z^2 = -x^2 -y^2$ but I'm not sure how to obtain $\rho$ from this.

I also tried $z=1$ but I'm not sure how to represent $z$ in spherical coordinates so that I can find $r$

The last one I tried is $y=\sqrt{1-x^2}$ and got $x^2 +y^2 =1$ but again, I'm not sure how to represent this in spherical coordinates.

I am able to do the actual integration once I have a triple integral with spherical coordinates set up, my issue is converting to spherical coordinates.

Overall I think my main issue is that I don't know any of the general conversions I should know in order to be able to answer this question, such as what $z$ is in spherical coordinates or what $x^2 +y^2$ is in spherical coordinates. If someone could show me how to convert to spherical coordinates it would help a lot.

(Also if anyone knows a website (or can be bothered to list them) where all the transformations I will need are listed please link it so I can try to learn these transformations so I can do these questions myself)

Thanks in advance

Answer 1

$ \displaystyle {\int_{-1}^1} {\int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}} {\int_{- \sqrt{x^2 + y^2}}^1} \ dz \ dy \ dx$

Note the region is bound by the following surfaces:

Cone $z = - \sqrt{x^2+y^2}$, cylinder $x^2+y^2 = 1$ and plane $z = 1$.

$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$

If we are integrating wrt $\rho$ first, we have to split the integral into two -

i) when $\rho$ is bound by the cylinder

$x^2 + y^2 \leq 1 \implies \rho \leq \csc\phi$

Using $z = - \sqrt{x^2+y^2}$, upper bound of $\phi = \cfrac{3\pi}{4}$

Lower bound of $\phi$ is at the intersection of cylinder and plane $z = 1$.

$z = 1 \implies \rho = \sec \phi$

So at intersection, $\sec\phi = \csc\phi \implies \phi = \cfrac{\pi}{4}$

ii) when $\rho$ is bound by the plane $z = 1$

$\rho \leq \sec\phi, 0 \leq \phi \leq \cfrac{\pi}{4}$

So integral is,

$\displaystyle \int_0^{2\pi} \int_{\pi/4}^{3 \pi/4} \int_0^{\csc\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta \ $ +

$\displaystyle \int_0^{2\pi} \int_{0}^{\pi/4} \int_0^{\sec\phi} \rho^2 \sin \phi \ d\rho \ d\phi \ d\theta$