Find the volume of the solid $Q$ cut from the sphere $x^2 + y^2 + z^2 ≤ 4$ by the cylinder $r = 2 \sin θ$ .

Question

I set up the integral as $$\int_0^{\pi}\int_0^{2\sin\theta}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dzdrd\theta$$ and got $\frac{16\pi}3$, but the answer is $\frac{16(3\pi-4)}9 \approx 9.64$, which is also what my calculator got too. What am I doing wrong?

$$\begin{split} \int_0^{\pi}\int_0^{2\sin\theta}\int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r dzdrd\theta &= \int_0^{\pi}\int_0^{2\sin\theta} r(2\sqrt{4-r^2})drd\theta \\ &= -\frac23\int_0^{\pi} (8\cos^3\theta - 8)d\theta \\ &= \frac{16}3\int_0^\pi (1-\cos^3{\theta})d\theta \\ &= \frac{16\pi}3 \end{split}$$

Answer 1

As stated in a reply above, the OP's mistake was not accounting for $\cos^3\theta$ being negative in the domain ${\pi\over2}$ to $\pi$. @JoséCarlosSantos was correct in splitting the bounds, changing $\frac{16}3\int_0^{\pi}1-\left(\cos^2\theta\right)^{3/2}\mathrm d\theta$ to $\frac{32}3\int_0^{\pi\over2}(1-\cos^3{\theta})\mathrm d\theta$, but gave a slightly incorrect final answer.

I will provide the work for solving this integral, in case anyone is interested: $$\frac{32}3\int_0^{\pi/2}1-\cos^3\theta\,\mathrm d\theta$$ Using using a handy cube identity, this can be rewritten as \begin{align*}\frac{32}3\int_{0}^{\pi/2}\left[1-\left(\frac14\cos3\theta+\frac34\cos\theta\right)\right]\mathrm d\theta&=\frac{32}3\left[\left(\left.\theta-\frac34\sin\theta-\frac1{12}\sin3\theta\right)\right|_{0}^{\pi/2}\right] \\&=\frac{32}3\left[\frac{\pi}{2}-\frac34+\frac1{12}\right] \\&=\frac{32\pi}6-\frac{64}9 \\&=\frac{16(3\pi-4)}9.\end{align*}

Answer 2

You have no justification to write that$$\int_0^{\pi}\int_0^{2\sin\theta} r(2\sqrt{4-r^2})\,\mathrm dr\,\mathrm d\theta = -\frac23\int_0^{\pi} 8\cos^3\theta - 8\,\mathrm d\theta.$$In fact,\begin{align}\int_0^{\pi}\int_0^{2\sin\theta} r(2\sqrt{4-r^2})\,\mathrm dr\,\mathrm d\theta&=\int_0^\pi\frac{16}3-\frac23\left(4-4\sin^2\theta\right)^{3/2}\,\mathrm d\theta\\&=\frac{16}3\int_0^\pi1-\left(\cos^2\theta\right)^{3/2}\,\mathrm d\theta\\&=\frac{32}3\int_0^{\pi/2}1-\cos^3\theta\,\mathrm d\theta\\&=\frac{16\pi}3-\frac{32}9.\end{align}