Integrating the Paraboloid: Finding y^2z^2

[yitriana]

Homework Statement

Find $$\int\int\int y^2 z^2$$where 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.
$$\int_{z = ?}^{?} \int_{\theta = ?}^{?} \int_{r = ?}^{?} <integrand> r dr d\theta dz$$

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]

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

 

[Mark44]

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