The equation of the cylinder is

$$x²+y²=ay \implies x²+\left(y-\frac{a}{2}\right)^2 = \frac{a²}{4} \tag{1}$$

and the sphere is $$x²+y²+z²=a²$$

In order to find this integral, I first realized that the region is a sphere inside a cylinder delocated from the center. I first computed the integral with respect to $z$ of the length of the sphere, which is:

$$\int_{-\sqrt{a-x²-y²}}^{\sqrt{a-x²-y²}}dz = 2\sqrt{a-x²-y²}$$

Now, I need to take this integral over the circumference that the cylinder forms when projected to the $xy$ plane, which is $(1)$. Therefore, I need to integrate $2\sqrt{a-x²-y²}$ inside the circumference $(1)$. I'll do it in polar coordinates, so I need to make my substitutions. The idea is to substitute $x=p\cos(t), y=p\sin(t)$ and not $x=p\cos(t), y-\frac{a}{2}=p\sin(t)$ because I want to make the integrand easier, not the region. So the region will not be simply $p$ from 0 to $\frac{a}{2}$ and $\theta$ from $0$ to $2\pi$. If I make the substition $x=p\cos(t), y=p\sin(t)$ in $(1)$ I end up with the equation $p=\sin(\theta)$ for the circumference $(1)$in polar coordinates. I also know that $\theta$ must go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ for this one. Therefore, my double integral becomes:

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{\sin(t)} 2\sqrt{a-(p\cos(\theta))²-(p\sin(\theta))²}p\ dp \ d\theta$$

Since this integral is quite hard, I would like to know if I'm doing all fine before trying to integrate this. Wolfram Alpha does not integrate it, don't know why, but my book says the answer is $$\frac{2\cdot a³\pi}{3}$$


$$\int_o^{\pi} (-1)\int_0^{a\sin(t)}\sqrt(a²-p²)-2p \ dp\ dt$$

$$u = a²-p² \implies du = -2p dp $$

$$\int_o^{\pi} (-1)\int_{a²}^{a²\cos²(t)}\sqrt(u)du\ dt=$$ $$\int_o^{\pi} (-1)\frac{2}{3}((a²\cos²(t))^{3/2}-(a²)^{3/2})dt = $$ $$(-1)\frac{2}{3}a^3\int_0^{\pi} (\cos^3(t)-1) dt = \frac{2\cdot a³\pi}{3}$$



First, the boundary of the sphere is $z = \pm\sqrt{a^\color{red}{2}-x^2-y^2}$ (note the exponent on $a$).

Also, the region bounded by $x^2+y^2 = ay$ is a circle of radius $\tfrac{a}{2}$ centered at $(0,\tfrac{a}{2})$. If you sketch this, you'll see that the bounds for $\theta$ are $0 \le \theta \le \pi$ instead of $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$.

So, your double integral (in polar coordinates) is $$\displaystyle\int_{0}^{\pi}\int_{0}^{a\sin \theta}2\sqrt{a^2-(p \cos \theta)^2-(p \sin \theta)^2}p\,dp\,d\theta$$

Using the identity $(\cos\theta)^2+(\sin\theta)^2 = 1$, you can simplify the integral as $$\displaystyle\int_{0}^{\pi}\int_{0}^{a\sin \theta}2\sqrt{a^2-p^2}p\,dp\,d\theta$$

For the inner integral, substitute $u = a^2-p^2$, $du = -2p\,dp$. This will give you something which is easy to integrate.

  • $\begingroup$ why from $0$ to $\pi$? For me, $-\pi/2$ to $\pi/2$ works: wolframalpha.com/input/… $\endgroup$ – Guerlando OCs Dec 14 '15 at 4:06
  • $\begingroup$ The circle $x^2+y^2=ay$ lies in the first and second quadrants. So if you substitute $x = p\cos\theta$ and $y = p\sin\theta$, then your range of $\theta$ will be $0$ to $\pi$ instead of $-\pi/2$ to $\pi/2$ (which goes over the fourth and first quadrants). For this problem, you might get the same answer using either set of bounds due to symmetry. $\endgroup$ – JimmyK4542 Dec 14 '15 at 4:16
  • $\begingroup$ thanks, I updated my question with the solution $\endgroup$ – Guerlando OCs Dec 14 '15 at 15:55

I think, conversely, that the double integrals are easy to handle. Note that we are facing a symmetric region so we can solve just: $$2\int_0^{\theta=\pi/2}\int_0^{a\sin(\theta)}\sqrt{a^2-r^2}~rdrd\theta$$


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