[MultiVarCalc] Find the mass of the solid

[bornofflame]

Homework Statement

Let $a > 0$. Find the mass of the "solid bowl" consisting of points inside the paraboloid $z=a(x^2+y^2) \text { for } 0\leq z \leq H \text{. Assume a mass density of } \rho(x, y, z) = z$.

Homework Equations

$x^2 + y^2 = z^2$

The Attempt at a Solution

[/B]
mass = $m =\int_W \rho(x, y, z)\, dV = \int \int \int \rho(x,y,z) dV = \int \int \int zr\, dz\, dr\, d\theta$
boundaries: $0 \leq z \leq H;~ 0 \leq r \leq \sqrt \frac H a;~ 0 \leq \theta \leq 2\pi$
$m = \int_0^{2\pi} \int_0^{\sqrt \frac H a} \int_0^H zr\,dz\,dr\,d\theta$
$= \int_0^{2\pi} \int_0^{\sqrt \frac H a} \frac 1 2 z^2r\left. \right|_0^H \,dr \,d\theta$
$=\frac 1 2 \int_0^{2\pi} \int_0^{\sqrt \frac H a} H^2r \,dr \,d\theta$
$= \frac 1 2 \int_0^{2\pi} \frac 1 2 H^2r^2\left. \right|_0^{\sqrt \frac H a}\, d\theta$
$= \frac 1 4 \int_0^{2\pi}| H^2(\sqrt \frac H a)^2\, d\theta$
$= \frac 1 4 \int_0^{2\pi} H^2(\frac H a)\, d\theta$
$= \frac 1 4 \int_0^{2\pi} \frac {H^3} a \,d\theta$
$= \frac 1 4 \frac {H^3} a \theta \left. \right|_0^{2\pi}$
$= \frac {H^3} {4a} 2\pi$
$= \frac {H^3} {2a}$

 

[LCKurtz]

Homework Statement

Let $a > 0$. Find the mass of the "solid bowl" consisting of points inside the paraboloid $z=a(\sqrt{x^2+y^2)} \text { for } 0\leq z \leq H \text{. Assume a mass density of } \rho(x, y, z) = z$.

That equation isn't a paraboloid. It is a cone. Which do you want?

Homework Equations

$x^2 + y^2 = z^2$

The Attempt at a Solution

View attachment 225683 [/B]
mass = $m =\int_W \rho(x, y, z)\, dV = \int \int \int \rho(x,y,z) dV = \int \int \int zr\, dz\, dr\, d\theta$
boundaries: $0 \leq z \leq H;~ 0 \leq r \leq \sqrt \frac H a;~ 0 \leq \theta \leq 2\pi$
$m = \int_0^{2\pi} \int_0^{\sqrt \frac H a} \int_h^H zr\,dz\,dr\,d\theta$

What is $h$? And in the step below you change it to $0$. That isn't correct in any case.

$= \int_0^{2\pi} \int_0^{\sqrt \frac H a} \frac 1 2 z^2r\left. \right|_0^H \,dr \,d\theta$
$=\frac 1 2 \int_0^{2\pi} \int_0^{\sqrt \frac H a} H^2r \,dr \,d\theta$
$= \frac 1 2 \int_0^{2\pi} \frac 1 2 H^2r^2\left. \right|_0^{\sqrt \frac H a}\, d\theta$
$= \frac 1 4 \int_0^{2\pi}| H^2(\sqrt \frac H a)^2\, d\theta$
$= \frac 1 4 \int_0^{2\pi} H^2(\frac H a)\, d\theta$
$= \frac 1 4 \int_0^{2\pi} \frac {H^3} a \,d\theta$
$= \frac 1 4 \frac {H^3} a \theta \left. \right|_0^{2\pi}$
$= \frac {H^3} {4a} 2\pi$
$= \frac {H^3} {2a}$