Spherical coordinates triple integral, help

Question

I have the following triple integral with the instructions

By using spherical coordinates, evaluate the integral: $$L = \iiint_{D} \frac{1}{\sqrt(x^2+y^2+z^2)} dxdydz$$

Where D is the portion of the unit cylinder $x^2 + y^2 \leq 1$ which lies between $z =0 $ and $z = 1$

My thoughts were that we use the bounds and use the Jacobian:

$$0 \leq \rho \leq \sqrt2, $$ $$0 \leq \theta \leq 2 \pi $$ $$0 \leq \varphi \leq \frac{\pi}{4}$$

To which after computing, we end up with the result $$\pi(\sqrt2 - 2)$$

I was hoping someone could either correct me, or verify. And if you do verify it, is there a method which is easier or different, because I would want to learn it thank you!

Answer 1

The region you are integrating over is cylinder $x^2 + y^2 \leq 1, 0 \leq z \leq 1$

So it is easier to find the limits of integration in cylindrical coordinates.

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

Your integrand becomes $\frac{1}{\sqrt{x^2 + y^2 + z^2}} = \frac{1}{\sqrt{r^2 + z^2}}$

And your integral is $\displaystyle \int_0^{2\pi}\int_0^1 \int_0^1 \frac{r} {\sqrt{r^2+z^2}} \, dr \, dz \, d\theta$

EDIT: In spherical coordinates,

Your integral is $\displaystyle \int_0^{2\pi} \int_0^{\pi/4} \int_0^{1 / \cos \phi} \rho \sin \phi\, d \rho \, d\phi \, d\theta + \int_0^{2\pi} \int_{\pi/4}^{\pi/2} \int_0^{1 / \sin \phi} \rho \sin \phi\, d \rho \, d\phi \, d\theta$

Here is a diagram that shows why $\rho$ is $\frac{1}{\cos \phi}$ (as $z \leq 1$) for $0 \leq \phi \leq \frac{\pi}{4}$.

Also why $\rho$ is $\frac{1}{\cos (\frac{\pi}{2} - \phi)} = \frac{1}{\sin \phi}$ (as radius of cylinder is $1$) for $\frac{\pi}{4} \leq \phi \leq \frac{\pi}{2}$.

Answer 2

That answer that you got is negative, and therefore it cannot possibly be correct.

You can do it in cylindrical coordinates:\begin{align}\iiint_D\frac{\mathrm dx\,\mathrm dy\,\mathrm dz}{\sqrt{x^2+y^2+z^2}}&=\int_0^{2\pi}\int_0^1\int_0^1\frac\rho{\sqrt{\rho^2+z^2}}\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta\\&=2\pi\int_0^1\sqrt{1+z^2}-z\,\mathrm dz\\&=\pi\left(\sqrt2-1+\operatorname{arcsinh}(1)\right).\end{align}

Answer 3

Your bounds aren't right. To use spherical coordinates, you'll have to breach the region into two pieces. One piece has bounds:

$$0\leq \rho \leq \sec\phi,\; 0\leq \phi \leq \pi/4,\; 0\leq \theta \leq 2\pi.$$

The other has:

$$0\leq \rho \leq \csc\phi,\; \pi/4\leq \phi \leq \pi/2,\; 0\leq \theta \leq 2\pi.$$

The equations $\rho = \sec\phi$ and $\rho = \csc\phi$ are the equations of horizontal and vertical lines in the $\rho\phi$-plane.