- #1

- 42

- 3

## Homework Statement

Given

[tex]

E = [(x,y,z) s. t. 0 \leq x \leq 2, 0 \leq y \leq \sqrt{2x - x^2}, 0 \leq z \leq 2]

[/tex]

Calculate

[tex]

\int_E z^3\sqrt{x^2+y^2}dxdydz

[/tex]

## Homework Equations

In cylindrical coordinates:

[tex]

x=rcos(\theta)\\y=rsin(\theta)\\z=z\\dxdydz = \rho d\rho d\theta dz

[/tex]

## The Attempt at a Solution

The integral is kind of weird to compute in Cartesian coordinates, so I tried to find what kind of curve are x and y bounded to vary on.

[tex]

y<\sqrt{2x-x^2}\\ y^2+x^2-2x<0

[/tex]

and by completing the square I got:

[tex]

y^2 + (x-1)^2 < 1

[/tex]

which is a circle of radius 1 centered in (1,0). So x and y vary in this circle and, given the upper bounds of x and z, I think this is the left half of a cylinder centered in ##(1,0)## with height 2.

Here comes my problem. I thought of using cylindrical coordinates to compute the integral and this leads to:

[tex]

x=\rho cos(\theta)+1\\y=\rho sin(\theta)\\z=z\\

0 \leq \rho \leq 1,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2},0 \leq z \leq 2

[/tex]

After plugging in the integral:

[tex]

\int^1_0 \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\int^2_0 z^3 \sqrt{\rho^2+1+2\rho cos(\theta)}\rho d\rho d\theta dz

[/tex]

Computing this integral for z is easy, but when it comes to ##\theta## and ##\rho## I have no idea how to solve it. Can anyone give me any suggestions?