Find the volume of the body that is limited by the plane given in the spherical coordinates

Question

Find the volume of the body that is limited by the plane given in the spherical coordinates at $$r=4-2\cos\varphi$$ where $\varphi$ is the azimuth.

I don't see what they are asking for in this question. How do I do it, and what answer do we get?

Answer 1

It turns out that the cross-section of the solid at a constant azimuth $\varphi$ is a semicircle, since the polar angle $\theta$ takes no part in the equation. This means that we can integrate to find the volume: $$\int_0^{2\pi}\int_0^\pi\int_0^{4-2\cos\varphi}1\cdot r^2\sin\theta\,dr\,d\theta\,d\varphi$$ $$=\int_0^{2\pi}\int_0^\pi\frac{(4-2\cos\varphi)^3}3\sin\theta\,d\theta\,d\varphi$$ $$=\int_0^{2\pi}\frac{2(4-2\cos\varphi)^3}3\,d\varphi$$ $$=\frac{16}3\int_0^{2\pi}(2-\cos\varphi)^3\,d\varphi$$ $$=\frac{16}3\int_0^{2\pi}(8-12\cos \varphi+6\cos^2\varphi-\cos^3\varphi)\,d\varphi$$ $$=\frac{16}3\int_0^{2\pi}\left(8-12\cos\varphi+3(1+\cos2\varphi)-\frac14(3\cos\varphi+\cos3\varphi)\right)\,d\varphi$$ $$=\frac{16}3\int_0^{2\pi}\left(11-\frac{51}4\cos\varphi+3\cos2\varphi-\frac14\cos3\varphi\right)\,d\varphi$$ $$=\frac{16}3\left[11\varphi-\frac{51}4\sin\varphi+\frac32\sin2\varphi-\frac1{12}\sin3\varphi\right]_0^{2\pi}=\frac{16}3(22\pi)=\frac{352\pi}3$$