I need to find the volume of the finite region enclosed between the surface
$$ y = 1 – x^2 – 4z^2 $$ and the plane $$y = 0$$
Here's what I've done:
$$ \int\int \left(\int_0^{1-x^2-4z^2}\mathrm{d}y)\right)\mathrm{d}A$$
where dA is the ellipse that the paraboloid projects down to in the x-z plane. For this ellipse, I've used the parametrisation
$$x = r\cos{\theta}, z = \dfrac{1}{2}r\sin\theta$$
and calculated the area in polar coordinates. Is this method legit?
Thanks
Best Answer
Basically you're trying: $$\iint_A y \,\mathrm{d}A = 2 \int_{0}^{1}\!\!\int_{0}^{2\pi} (1-r^2) \cdot r \,\mathrm{d}\theta \,\mathrm{d} r$$
The scale factor of 2 is because without it the volume would be for between $y_2=1-x^2-z^2$ and the $x,z-$plane. (You applied a scaled polar transformation to $z$).