# How do you find the area of the solid enclosed by the graphs z=25(x^2+y^2) and z=8?

Feb 15, 2015

The answer is: $V = \frac{32}{25} \pi$.

this function is a circular paraboloid, with the center in the origin of axes, and it is concave up.

The volume requested can be calculated with this triple integral:

$\int \int {\int}_{D} 1 \mathrm{dx} \mathrm{dy} \mathrm{dz}$,

where $D$ is the domain, in this case the function:

$z = 25 \left({x}^{2} + {y}^{2}\right)$.

The explanation is that this integral:

$\int \int {\int}_{D} f \left(x , y , z\right) \mathrm{dx} \mathrm{dy} \mathrm{dz}$

is a 4-dimension volume with "basis" $D$ and "height" $f \left(x , y , z\right)$. Since $D$ is a volume, if we put the function $1$ as height, we obtain the volume $D$ itself!

For this calculus we have to change the coordinate system. Instead of the cartesian coordinates, we have to use the cylindrical polar coordinates, that are:

$x = \rho \cos \theta$
$y = \rho \sin \theta$
$z = z$

and we have to remember that we have to calculate the Jacobian Matrix of the passing from a system of coordinates to an other, that is:

$\mathrm{dx} \mathrm{dy} \mathrm{dz} = \rho \mathrm{dr} h o d \theta \mathrm{dz}$.

Since $z = 25 \left({x}^{2} + {y}^{2}\right)$ and ${x}^{2} + {y}^{2} = {\rho}^{2} {\cos}^{2} \theta + {\rho}^{2} \sin \theta =$

$= {\rho}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = {\rho}^{2} \cdot 1 = {\rho}^{2}$,

our function becomes:

$z = 25 {\rho}^{2}$.

We have to find the volume under the plane $z = 8$ so $0 \le z \le 8$.

Since $\rho$ has to be limited from the function, ${\rho}^{2} \le \frac{z}{25} \Rightarrow | \rho | \le \frac{\sqrt{z}}{5} \Rightarrow$

$- \frac{\sqrt{z}}{5} \le \rho \le \frac{\sqrt{z}}{5}$,

but $\rho \ge 0$, so $0 \le \rho \le \frac{\sqrt{z}}{5}$.

The volume is in every part of every circle, sections of $D$ with orizontal planes, so:

$0 \le \theta \le 2 \pi$.

Finally the integral becomes:

$V = {\int}_{0}^{2 \pi} {\int}_{0}^{8} {\int}_{0}^{\frac{\sqrt{z}}{5}} \rho \mathrm{dr} h o d \theta \mathrm{dz} =$

$= {\int}_{0}^{2 \pi} d \theta {\int}_{0}^{8} {\left[{\rho}^{2} / 2\right]}_{0}^{\frac{\sqrt{z}}{5}} \mathrm{dz} =$

$= {\left[\theta\right]}_{0}^{2 \pi} {\int}_{0}^{8} \frac{\frac{z}{25}}{2} \mathrm{dz} = 2 \pi \frac{1}{50} {\left[{z}^{2} / 2\right]}_{0}^{8} =$

$= \frac{\pi}{25} \left(\frac{64}{2}\right) = \frac{32}{25} \pi$.