I need a check on my solutions - I don't have answers in my study guide.

Also would appreciate commens on my 'train of thought' ie if there is a shorter way to arrive at my findings.

2. Find the volume under the parabolic surface $\displaystyle z=x^2+y^2$ for x, y such that $\displaystyle x^2+y^2\leq{a^2}$.

Answer.

Define the region of integration:

$\displaystyle x^2+y^2\leq{a^2}$from there $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{a^2}\leq1$. Therefore, the area under the function z is the ellipse with identical foci (a=b) ie cirlce.

Since there is no one function that describes the circle, I can either split it into two parts and evaluate two integrals or - which I'd rather prefer - observe that the two areas of integration are symmetrical and are described by the functions symmetrical with relation to both X or Y axis, and only evaluate one integral, then double its value to get to the required volume.

Is my short-cut strategy valid?...

To evaluate one half part (one double integral), I will use variable substitution $\displaystyle x=rcos\Theta y=rsin\Theta$

$\displaystyle \int_0^{\pi}d\Theta\int_0^ar^2\frac{drd\Theta}{r}= \int_0^{\pi}\frac{a^2}{2}d\Theta=\frac{\pi*a^2}{2}$

I will need to double that to find the total volume which is $\displaystyle \pi*a^2$. Looking back at the solution, however, something does not seem right as this looks too much like an area, not a volume...