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Find the mass of the lamina bounded by the curve $ay^2=x^3$ and the line $by = x$ if the density at a point varies as the distance of the point from the x-axis.
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The point of interaction is $ay^2 = x^3, by = x$

$ay^2 = (by)^3$

$b^3y^3 - ay^2 = 0 \Rightarrow y^2(b^3y-a)=0$

$y = \frac{a}{b^3}, x = \frac{a}{b^2}$

$\therefore$ the point of intersection is $\Big(\frac{a}{b^2}, \frac{a}{b^3}\Big)$

The lamina is the area OBA on the curve OBA $y=\frac{x^{3/2}}{\sqrt{a}}$ and on the line OA, $y = \frac{x}{b}$

The surface density is given by $\rho = ky$ taking the strip parallel to y - axis.

Max of the lamina $= k\int_0^{a/b^2} \int_{\frac{x^{3/2}}{\sqrt{a}}}^{x/b} y \hspace{0.1cm}dx \hspace{0.1cm}dy$

$\hspace{2.8cm}= k \int_0^{a/b^2}\Big[\frac{y^2}{2}\Big]^{x/b}_{\frac{x^{3/2}}{\sqrt{a}}}dx\\ \hspace{2.7cm}=\frac{k}{2}\int^{a/b^2}_0\Big(\frac{x^2}{b^2}-\frac{x^3}{a}\Big)dx\\ \hspace{2.7cm} = \frac{k}{2}\Big[\frac{x^3}{3b^2}-\frac{x^4}{4a}\Big]^{a/b^2}_0\\ \hspace{2.7cm} = \frac{k}{2}\Big[\frac{a^3}{3b^8}-\frac{a^3}{4b^8}\Big]\\ \hspace{2.7cm} = \frac{k}{2}\frac{a^3}{b^8}\Big[\frac{1}{3} - \frac{1}{4}\Big]\\ \hspace{2.7cm} = \frac{k}{24}\frac{a^3}{b^8} $

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