Question

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

Answer 1

The answer is: `V=32/25pi`.

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:

`intintint_D1dxdydz`,

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

`z=25(x^2+y^2)`.

The explanation is that this integral:

`intintint_Df(x,y,z)dxdydz`

is a 4-dimension volume with "basis" `D` and "height" `f(x,y,z)`. 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=rhocostheta`
`y=rhosintheta`
`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:

`dxdydz=rhodrhod thetadz`.

Since `z=25(x^2+y^2)` and `x^2+y^2=rho^2cos^2theta+rho^2sintheta=`

`=rho^2(cos^2theta+sin^2theta)=rho^2*1=rho^2`,

our function becomes:

`z=25rho^2`.

We have to find the volume under the plane `z=8` so `0<=z<=8`.

Since `rho` has to be limited from the function, `rho^2<=z/25rArr|rho|<=sqrtz/5rArr`

`-sqrtz/5<=rho<=sqrtz/5`,

but `rho>=0`, so `0<=rho<=sqrtz/5`.

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

`0<=theta<=2pi`.

Finally the integral becomes:

`V=int_0^(2pi)int_0^8int_0^(sqrtz/5)rhodrhod thetadz=`

`=int_0^(2pi)d thetaint_0^8[rho^2/2]_0^(sqrtz/5)dz=`

`=[theta]_0^(2pi)int_0^8(z/25)/2dz=2pi1/50[z^2/2]_0^8=`

`=pi/25(64/2)=32/25pi`.