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Find the mass of a lamina in the form of an eclipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1$, if the density at any point varies as the product of the distance from the azes of the ellipse.
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Solution:

Mass of lamina is given by, $\mathrm{M}=\iint r\ d x \ d y$

r is the density function $r=k \ x \ y$

Ellipse eqn is $: \frac{x^{2}}{g^{2}}+\frac{y^{2}}{b^{2}}=1$

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$\begin{array}{rl}{0} & { \leq y \leq b \sqrt{a^{2}-x^{2}} / a} \\ {0} & { \leq x \leq a} \\ {\therefore M=4 \int_{0}^{a} \int_{0}^{b \sqrt{a^{2}-x^{2}} / a}} & {k x \ y \ d y \ d x}\end{array}$

$=4 \mathrm{k} \int_{0}^{a} x \cdot\left[\frac{y^{2}}{2}\right]^{b \sqrt{a^{2}-x^{2}}/a}_0 dx$

$\begin{aligned}=& 2 \mathbf{k} \int_{0}^{a} x \cdot \frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right) d x \\ &=\frac{2 k b^{2}}{a^{2}} \int_{0}^{a}\left[a^{2} x-x^{3}\right] d x \end{aligned}$

$\begin{aligned} &=\frac{2 k b^{2}}{a^{2}}\left[\frac{a^{2} x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{a} \\ \therefore \mathbf{M}=\frac{k a^{2} b^{2}}{2} \end{aligned}$

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