[MultiVarCalc] Find the volume of the solid

[bornofflame]

Homework Statement

Find the volume of the solid between the cone $z = \sqrt{x^2 + y^2}$ and the paraboloid $z = 12 - x^2 - y^2$.

Homework Equations

$x^2 + y^2 = r^2$

The Attempt at a Solution

I drew a simple diagram to start off with to visualize the solid formed by the intersection of these equations which is in the attached image along with my work in raw form.

Since we're dealing with circular cross sections the polar coordinates will be best, (read easiest), to work with for me.

Volume = ?
Setting up the integral
$V = \int_R dV = \int_R r\, dz\, dr\, d\theta$
$z = \sqrt{x^2 + y^2} = \sqrt{r^2} = r$
$z = 12 - x^2 - y^2 = 12 - r^2$

lower limit for z $= r$
upper limit for z $= 12 - r^2$

$\sqrt{x^2 + y^2} = 12 - x^2 - y^2$
$r = 12 - r^2$
$r^2 + r - 12 = 0$
$(r + 5)(r - 3) = 0$
$r = -4, 3: -4 \text{ is out of bounds, so }r = 3$
$0 \leq r \leq 3$

$V = \int_0^{2\pi} \int_0^3 \int_r^{r^2} r\, dz\, dr\, d\theta$
$= \int_0^{2\pi} \int_0^3 rz\, \left. \right|_r^{r^2}\, dr\, d\theta$
$= \int_0^{2\pi} \int_0^3 r^3-r^2\,dr\,d\theta$
$=\int_0^{2\pi} \frac 1 4 r^4- \frac 1 3 r^3 \left. \right|_0^3 \,d\theta$
$=\int_0^{2\pi} \frac 81 4 - 9\, d/theta = \frac 81 4 \theta - 9\theta \left. \right|_0^{2\pi}$
$=\frac {81} 4 (2\pi)-9(2\pi) = \frac {81} 2\pi - 18\pi = \frac {81} 2\pi - \frac {36} 2 \pi$
$=\frac {45} 2 \pi$

 

[andrewkirk]

It looks like the upper limit in the innermost integral, the one over $z$, is wrong. It needs to be the $z$ vertical coordinate of the paraboloid where the horizontal coords are $(r,\theta)$. That will not be $r^2$.

 

[bornofflame]

Should the upper limit be $12 - r^2$?

Which would give me $\frac {99} 2 \pi$ as an answer instead.

 

[andrewkirk]

Should the upper limit be $12 - r^2$?

Yes