Integrating the Paraboloid: Finding y^2z^2

[yitriana]

Homework Statement

Find \int\int\int y^2 z^2where E is the region bounded by the paraboloid x = 1 - y²2 - z² and the plane x = 0.

The Attempt at a Solution

The region is a paraboloid with vertex at x = 1, y = 0, z = 0. I chose z bounds to be between 0 and 1 - y²2 - z² for first integral. Then, I realized that since region was bounded by x = 0 plane, the y and z values would range (in polar coordinates), from 0 to 2\pi for y (or z) and 0 to 1 for z.

Then, upon finishing first integral for z bounds, I got (1 - y² - z²) * y²*z², and when converting to polar coordinates, I got,

(1 - r²)*(r⁴*cos²(\theta)*sin²(\theta)

I don't know how to simplify this expression so that I can integrate for theta. How do I do it?

 

[Mark44]

Is there a typo in x = 1 - y²2 - z² or did you mean x = 1 - 2y² - z²?
The usual practice is to put numerical coefficients before variables.

 

[yitriana]

sorry, it is meant to read x = 1 - y2 - z2

 

[Mark44]

You seem to be ignoring the differentials dx, dy, and dz in your first integral, and the dr d\theta and dz in your integral converted to polar form.

Due to the symmetry of your region and the integrand, you can take \theta between 0 and \pi/2, and multiply the resulting integral by 4.

It would be helpful to see your integral with limits and with differentials. You can see my LaTeX code just by clicking it.
<br /> \int_{z = ?}^{?} \int_{\theta = ?}^{?} \int_{r = ?}^{?} &lt;integrand&gt; r dr d\theta dz<br />

You'll need to fill in the lower and upper limits of integration, and the integrand will need to be converted to polar form as well.

 

[yitriana]

<br /> <br /> \int_{r = 0}^{1} \int_{\theta = 0}^{2 \pi} (1 - r^2) r^4 (\cos(\theta))^2 (\sin(\theta))^2 d\theta dr <br /> <br />

 

[Mark44]

You have skipped a step. Let's start from the triple integral that I provided.