0
$\begingroup$

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 $$\frac{2\cdot a³\pi}{3}$$

UPDATE:

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

$$u = a²-p² \implies du = -2p dp $$

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

:)

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.