I need to find the area of the portion of the cone $x^2+y^2=z^2$ above the $xy$ plane and inside the cylinder $x^2+y^2=ax$ .

For this, I used cylindrical coordinates to parametrize the region:

$$x=r\cos \theta$$ $$y=r\sin \theta$$ $$z=z$$

Since $x^2+y^2=z^2$ then $r=z$. And since $x^2+y^2=ax$ then $r=a\cos \theta$. So the final parameterization, for $-\pi/2<\theta<\pi/2$ and $0<r<a\cos \theta$, is:

$$x=r\cos \theta$$ $$y=r\sin \theta$$ $$z=r$$

And so, $T_r =(\cos\theta,\sin\theta,1)$ and $ T_{\theta}=(-r\sin\theta ,r \cos\theta, 0)$. Which gives $T_r \times T_{\theta}=(-r\cos\theta, -rsin\theta, r^2)$

And so the area is given by:

$$\int_{-\pi/2}^{\pi/2} \int_0^{a \cos\theta}r(1+r^2)^{1/2} dr d\theta$$

Which seems simple, but gives a hell-difficult integral when integrating respect to $\theta$. So I wanted to ask you: is my approach correct? Is there a simpler way to parametrize this surface?


1 Answer 1


The final double-integral you have doesn't evaluate to an elementary function, but it can be recognized as a combination of Elliptic Integrals.

Take the inner integral first. By putting $u=1+r^2$, It is easy to show that $$ \int_{0}^{a\cos\theta} r\sqrt{1+r^2}dr = \frac{1}{3} \sqrt{a^2 \cos ^2\theta +1} \left(a^2 \cos ^2 \theta + 1 \right) - \frac{1}{3} $$ So, your outer integral is $$ \int_{-\pi/2}^{\pi/2} \frac{1}{3} \sqrt{a^2 \cos ^2\theta +1} \left(a^2 \cos ^2 \theta + 1 \right) - \frac{1}{3} d\theta $$

For $a \neq 1$, it is well known that one gets elliptic integrals of various kinds. The limits $-\pi/2, \pi/2$ suggest that we should be able to get by with complete elliptic integrals. From the definition of $K(k)$ and $E(k)$, we have $$ K(k) = \int_{0}^{\pi/2} \frac{1}{\sqrt{1-k^2 \sin^2 \theta}} d\theta \\ E(k) = \int_{0}^{\pi/2} \sqrt{1-k^2 \sin^2 \theta} d\theta $$ You can use these to evaluate the integral above. For instance, the even integral $$ \int_{-\pi/2}^{\pi/2} \sqrt{a^2 \cos ^2\theta +1} \; d\theta \\ = 2 \int_{0}^{\pi/2} \sqrt{1 + a^2 - a^2 \sin ^2\theta} \; d\theta \\ = 2\sqrt{1+a^2} \int_{0}^{\pi/2} \sqrt{1 - \frac{a^2}{1+a^2} \sin ^2\theta} \; d\theta \\ = 2\sqrt{1+a^2} E \left( \frac{a^2}{1+a^2} \right) $$ and so on.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .