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.


1 Answer 1


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


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