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{\bf Exam Question

Topic: Double Integral}

Let $R$ denote the region in the first quadrant $(x>0, y>0)$
bounded by the hyperbolas $xy=1,xy=9$ and the lines $y=x,y=4x$.

Use the change of variable given by $x=u/v, y=uv$ to evaluate the
double integral

$$\int \!\!\! \int_R \left(\sqrt{\frac{y}{x}}+\sqrt{xy}\right)\,
d(x,y)$$

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{\bf Solution}

$$x=\frac{u}{v}, y=uv \ \ \mathrm{so}\ \ xy=u^2, \frac{y}{x}=v^2$$
The Jacobian for this transformation is
$$\frac{\partial(x,y)}{\partial(u,v)}=\left|\begin{array}{cc}
\frac{1}{v} & -\frac{u}{v^2}\\ v & u \end{array}\right|$$

$$\mathrm{So}\ \int \!\!\! \int_R
\left(\sqrt{\frac{y}{x}}+\sqrt{xy}\right)\, d(x,y) =\int \!\!\!
\int_{R'}(v+u).\frac{2u}{v}\, d(u,v)$$ $$=\int_{u=1}^3\,
du\int_{v=1}^2 \left( 2u+\frac{2u^2}{v}\right)\, dv
=\int_{u=1}^3\left[2uv+2u^2\ln v\right]_{v=1}^2\, du$$
$$=\int_1^3(2u+2u^2\ln 2)\, du=\left[u^2+\frac{2}{3}u^3\ln
2\right]_1^3$$ $$=8+\frac{2}{3}\times 26\times\ln 2=20.0146\
\mathrm{(4\ d.p.)}$$

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