multiple integralmultivariable-calculus
I want to find the volume of the region of the sphere $x^2+y^2+z^2=1$, between the planes $z=1$ and $z=\frac{\sqrt{3}}{2}$
I have used triple integral for calculating this
$$\int _0^{2\pi }\int _{0}^{\frac{\pi }{6}}\int _{0 }^1\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$
Are the limits of integral i have chosen correct??
Edit:
As from the comment, my integral is not correct, so please clarify what region actually the above integral represents.
turns out that I was viewing the figures in the wrong way. As Andrew D. Hwang pointed out the cylinder has infinite height. Since the cylinder ends are open, it engulfs the top and bottom end of the sphere so the endcaps are not included. This explains why the integral:
$$\int _0^{2\pi }\int _{\frac{\pi }{6}}^{\frac{5\pi }{6}}\int _{csc\phi }^2\:\rho ^2sin\phi \:d\rho \:d\phi \:d\theta $$
is correct for this problem.
Using spherical coordinates as follows $$y=r\sin\theta\cos\phi$$ $$x=r\sin\theta\sin\phi$$ $$d\sigma=\text{Elementary surface area of sphere}$$$$=(r\sin\theta d\phi)(rd\theta)=r^2\sin\theta d\theta d\phi$$
Now, $$\int\int_S(x^2+y^2)d\sigma$$ $$=\int_{0}^{2\pi}\int_{0}^{\pi}(r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi)r^2\sin\theta d\theta d\phi$$ $$=\int_{0}^{2\pi}\int_{0}^{\pi}(r^2\sin^2\theta(\cos^2\phi+\sin^2\phi))r^2\sin\theta d\theta d\phi$$
$$=\int_{0}^{2\pi}\int_{0}^{\pi}r^4\sin^3\theta d\theta d\phi$$ $$=r^4\int_{0}^{2\pi}\int_{0}^{\pi}\sin^3\theta d\theta d\phi$$ Setting $r=\text{radius of unit sphere}=1$ $$=\int_{0}^{2\pi}\left[\int_{0}^{\pi}\frac{3\sin\theta-\sin 3\theta}{4}d\theta \right]d\phi$$ $$=\int_{0}^{2\pi}\left[-\frac{3}{4}\cos\theta+\frac{1}{12}\cos 3\theta \right]_{0}^{\pi}d\phi$$ $$=\int_{0}^{2\pi}\left[\frac{3}{2}-\frac{1}{6} \right]d\phi$$ $$=\int_{0}^{2\pi}\left[\frac{4}{3}\right]d\phi$$ $$=\frac{4}{3}\int_{0}^{2\pi}d\phi$$ $$=\frac{4}{3}[\phi]_{0}^{2\pi}$$ $$=\frac{4}{3}[2\pi]=\frac{8\pi}{3}$$
Best Answer
The bottom of your region is a horizontal plane which is not the $xy$-plane. This means you should at least consider cylindrical coordinates. For instance, $$ \int_0^{2\pi}\int_0^{1/2}\int_{\sqrt3/2}^{\sqrt{1-r^2}} r\,dz\,dr\,d\theta $$ or $$ \int_0^{2\pi}\int_{\sqrt3/2}^1\int_0^{\sqrt{1-z^2}}r\,dr\,dz\,d\theta $$ In case you really want spherical coordinates, if we put the radius as the innermost integral, note that it doesn't go from $0$, it goes from wherever $z=\sqrt3/2$, which is to say from $\frac{\sqrt3}{2\cos\phi}$, which is not easy to integrate. If we do $\phi$ first instead, we get $$ \int_0^{2\pi}\int_{\sqrt3/2}^1\int_0^{\arccos(2\rho/\sqrt3)}\rho^2\sin\phi\,d\phi\,d\rho\,d\theta $$ which might look scary at first, but everything turns out pretty nice in the end.