# Given f(x, y)=x^2+y^2-2x, how do you the volume of the solid bounded by z=(f(x, y)+f(y,x))/2-5/2, z = +-3?

Oct 26, 2016

$9 \pi$ cubic units.

#### Explanation:

The section of this sold by a plane parallel to the xy-plane is the

circle with

center at $\left(\frac{1}{2} , \frac{1}{2} , z\right)$and radius $R \left(z\right) = \sqrt{z + \frac{3}{2}}$

For integration to find the volume V, choose an element in the form

of a circular disc of thickness $\Delta z$ and radius R. The faces of

this disc are parallel to the xy-plane.

Now, V = limit $\Delta z \to 0$ of $\sum \pi {R}^{2} \Delta z = \int \pi {R}^{2} \mathrm{dz}$,

between the limits,from $z = - 3$ to z = 3..

So, $V = . \pi \int \left(z + \frac{3}{2}\right) \mathrm{dz}$, between the limits

$= \pi \left[{z}^{2} / 2 + \frac{3}{2} z\right] ,$ between $z = - 3 \mathmr{and} z = 3$

$= 9 \pi$ cubic units.