0
$\begingroup$

The given integral is: $$\int_{-7}^7\int_{-\sqrt{49-x^2}}^\sqrt{49-x^2}\int_{x^2+y^2}^{49}xdxdydx$$ Converting to cylindrical coordinates: $$\int_{0}^{2\pi}\int_{0}^7\int_{r^2}^{49}r^2\cos(\theta)dxdrd\theta$$ It's the spherical coordinates I'm having trouble converting too because I'm don't know how to sketch the given limits. I know that one of the equations is $x^2+y^2+z^2 = 49$ but because the limit of $\theta$ is $2\pi$ I am lost and the example my professor did only went to $\frac{\pi}{2}$. I can't find $\rho$ because I don't know where the two shapes intersect. I calculated the triple integral for cylindrical coordinates and got zero so I'm very confused.

$\endgroup$
0
1
$\begingroup$

If the integral is

$\displaystyle \int_{-7}^7\int_{-\sqrt{49-x^2}}^\sqrt{49-x^2}\int_{x^2+y^2}^{49} x \, dz \, dy \, dx$

We are basically integrating over the region bound between paraboloid $z = x^2 + y^2$ and plane $z = 49$.

It is a bit complicated in spherical coordinates and you have to split it into two regions.

$x = \rho \cos \theta \sin \phi$
$y = \rho \sin \theta \sin \phi$
$z = \rho \cos \phi$

In spherical coordinates, $\phi$ is the angle made with $z$ axis so at $z = 49$, $\phi$ varies from $0$ to $\cot \phi = \frac{49}{7} = 7 \implies \cot^{-1}(7)$ with no change in value of $z$.

Also as $z$ is constant, $\rho$ as a function of $\phi$ will be $0 \leq \rho \leq 49 \sec \phi$ (from equation of $z$).

Now for $\cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}$, we are traversing till the paraboloid boundary

So, $z = x^2 + y^2 \implies \rho \cos \phi = \rho^2 \sin^2 \phi \implies \rho = \cot \phi \csc \phi$

So, $I = \iiint \rho \cos \theta \sin \phi.\rho^2 \sin \phi \, dr d\phi d \theta$

with two integrations in two parts with limits of integration -

i) $0 \leq \rho \leq 49 \sec \phi, 0 \leq \phi \leq \cot^{-1}(7), 0 \leq \theta \leq 2\pi$.

ii) $0 \leq \rho \leq \cot \phi \csc \phi, \cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}, 0 \leq \theta \leq 2\pi$.

$\endgroup$

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.