# Triple integral in different coordinate systems.

My task is to write volume integral in 3 coordinate systems : Cartesian, cylindrical and spherical. This integral shows volume of intersection of 2 spheres, first with center at $(0, 0,-3)$ and radius $5$ and second with center at $(0,0, 3)$ and radius $\sqrt{13}$.I am able to do it for Cartesian and cylindrical systems:

$$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\int_{3-\sqrt{13-x^2-y^2}}^{-3+\sqrt{25-x^2-y^2}} \, dz\,dy\,dx,$$

Update: as it was pointed out by Ross Millikan, cylindrical version should looks like this: $$\int_{0}^{2\pi}\int_{3-\sqrt{13}}^{1}\int_{0}^{\sqrt{13-(z-3)^2}}r \, dz\,dr\,d\varphi + \int_{0}^{2\pi}\int_{1}^{2}\int_{0}^{\sqrt{25-(z+3)^2}}r \, dz\,dr\,d\varphi.$$

However, I can not handle spherical case. Please help me to understand it. Please show me how to make substitution. What I need is understanding how to solve this problem, since I need to solve more than problem of this type.

Thanks a lot for your help!

Update:

Now I am stuck with understanding the first term of spherical part. My attempt:

$$\int_{0}^{2\pi}\int_{0}^{3}\int_{\pi}^{\arctan\frac{r}{3-\sqrt{13-r^2}}}r^2 \cos \psi \, d \psi \,dr\, d \varphi \, .$$ But numerically it is different from first cylindrical term.

Neither of your integrals is correct. You are integrating over the intersections of two balls. Spheres are the boundary of the balls. You need to find the $z$ coordinate where the spheres intersect, call it $z'$, then integrate over the top ball up to that $z$ position and integrate over the bottom ball above it. The $\varphi$ integral should be $0$ to $2\pi$ as you have it. The lowest point of the top ball is at $z=3-\sqrt {13}$ and the highest point of the bottom ball is at $z=2$. This gives $$\int_0^{2\pi}\int_{3-\sqrt{13}}^{z'}\int_0^{\sqrt{13-(z-3)^2}}rdr\ dz\ d\varphi$$ plus another term from $z'$ to $2$ in $z$
• Thanks a lot for the answer! Can you please show me first term in spherical coordinates? $z'=1$. I will compute both terms for both cylindrical and spherical coordinates and compare them asap. Thanks a lot for pointing out on my mistake! May 21, 2017 at 21:58
• For the spherical one, the top part of the bottom ball extends from $\theta=0$ to $\theta=\arccos \frac 1{\sqrt{13}}$ and you can use the law of cosines to find $r(\theta)$ over that range. If you draw the triangle from the lower center to the origin to a point on the edge of the ball you get $5^2=3^2+r^2-2\cdot 3 \cdot r \cos(\pi-\theta)$. Solve that for $r$ and plug into your integral. The Cartesian one needs a fix similar to the cylindrical one. Then add in a similar expression for the lower part. May 21, 2017 at 23:07
• The $\varphi$ integral does not interact with the others, so it can be done any time. If you get $r(\theta)$ for the sphere, you integrate from $0$ to that $r(\theta)$ then take $\theta$ from $0$ to $\arccos \frac 1{\sqrt{13}}$. Then do another integral with $\theta$ from $\arccos \frac 1{\sqrt{13}}$ to $\pi$ with the other $r(\theta)$ May 21, 2017 at 23:18
I'm not quite sure what your spherical coordinate $$\psi$$ is, as I'm accustomed to measuring the angle from the positive $$z$$-axis and having the formula $$dV = r^2\sin\psi\,dr\,d\psi\,d\phi$$. So I'm guessing you're measuring $$\psi$$ to be the angle with the $$z$$-axis measured with $$\psi=0$$ at the $$xy$$-plane.
Regardless, you need to find at what angle $$\psi$$ the two spheres intersect and then break the integral into two pieces. The awkward thing is that the equation for a sphere centered away from the origin is somewhat yucky unless the sphere happens to pass through the origin (and neither of yours does). You can substitute into the cartesian equations to find that the spheres have spherical equations $$r^2 + 6r\sin\psi = 16 \qquad\text{and}\qquad r^2 - 6r\sin\psi = 4,$$ respectively. Thus, they intersect when $$r=\sqrt{10}$$ and $$r\sin\psi = 1$$, so $$\psi = \arcsin(1/\sqrt{10})$$. This gives us the integrals $$\int_0^{2\pi}\int_{-\pi/2}^{\arcsin(1/\sqrt{10})}\int_0^{r_1(\psi)} r^2\cos\psi\,dr\,d\psi\,d\theta + \int_0^{2\pi}\int_{\arcsin(1/\sqrt{10})}^{\pi/2}\int_0^{r_2(\psi)} r^2\cos\psi\,dr\,d\psi\,d\theta,$$ where $$r_1(\psi) = 3\sin\psi+\sqrt{9\sin^2\psi+4}$$ and $$r_2(\psi) = -3\sin\psi+\sqrt{9\sin^2\psi+16}$$. I was wrong. This is very yucky, not somewhat yucky.