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bornofflame

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## 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

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}##

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