The solid sphere $\{ (x,y,z)\in R^3 ; x^2+y^2+z^2\le 1\}$ is cut into two parts by the plane $z=\frac{1}{2}$. I’m required to find the volume of the smaller part. I know how to do double and triple integrals essentially. But I have difficulty visualising problems like this. How do I proceed here?
Edit: I also want to do similar problems, so if anyone can link me to an article/site explaining relevant theory,do post it.
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It this case you can observe that you want to compute the volume of a solid of revolution, which is in given by $$ \int_{1/2}^{1} \pi (\sqrt{1-z^2})^2 du = \pi[z-\frac 12 z^2]_{1/2}^{1}=\frac{5 \pi}{24}. $$
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The volume is a spherical cap with the base circle given by $ x^2+y^2 = 1-\frac1{2^2} = \frac34$. Then, integrate the volume in cylindrical coordinates with $z(r) = \sqrt{1-r^2}$ and the limits $r \le\frac{\sqrt3}2$ and $0\le \theta \le 2\pi$.
$$\int_0^{2\pi}\int_0^{\sqrt3/2} (z(r)-\frac12)rdrd\theta =2\pi\int_0^{\sqrt3/2} (\sqrt{1-r^2}-\frac12)rdr=\frac{5\pi}{24}$$
Here is an image of the whole situation:
The $xy$-plane is in gray, the $z = \frac12$ plane is in blue. We want the volume of the top part.
There are several ways to do this. Mainly we need to choose a coordinate system. I'll go with Cartesian, because of its familiarity, but it's probably actually most convenient to do this in cylindrical coordinates (e.g. see it as a solid of rotation about the $z$-axis). Spherical coordinates are made slightly tricky by the $z = \frac12$ plane, but that's not entirely impossible either.
(Most of the time, choosing a coordinate system so that the relevant region is nicely presentable more than outweighs the possibly more inconvenient form of the volume element. Respect the symmetry you're given, and you are generally rewarded. In this case, it is simple enough for me to refrain, though.)
So, in Cartesian coordinates, I believe the most convenient way is to do $z$-integral outermost. The plane makes $z = \frac12$ a very nice and natural lower bound for the entire thing, and the upper bound must be $z = 1$. So, what needs integrating with respect to $z$ here? Since we're after a volume, integrating a function $A(z)$ that yields an area for each value of $z$ is very natural.
And the most natural area for any given $z$-value is the cross-sectional area of the sphere, which is to say, the area of a certain circle. Imagine a plane in my picture, parallel to the blue one, moving from the blue one up to the top of the sphere. The cross section between that and the sphere is what we're integrating.
This gives $$ V_{\text{upper}} = \int_{1/2}^1A(z)\,dz $$ So, what is this area function? Well, for any given $z$ value, the radius of the cross-section circle is $\sqrt{1-z^2}$. And the area of a circle with that radius is $\pi(1-z^2)$. So that's what we get: $$ V_{\text{upper}} = \int_{1/2}^1\pi(1-z^2)\,dz = \pi\left(1 - \frac13\right) - \pi\left(\frac12 - \frac1{24}\right)\\ = \frac{5\pi}{24} $$