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{\bf Exam Question

Topic: Double Integral}

By changing the variables so that the region $R$ is transformed
into a square, evaluate the integral $$\int \!\!\!
\int_R\left(5(y-x^2)^4+6xy\right)\left(y+2x^2\right)\, d(x,y),$$
where $R$ is the region in the first quadrant bounded by the
curves $$y=x^2,\ \ y=x^2+1,\ \ xy=1,\ \ xy=2.$$

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{\bf Solution}

Let $u=y-x^2;\ \ v=xy.$
$$\left|\frac{\partial(u,v)}{\partial(x,y)}\right|=\left|\left(\begin{array}{cc}
-2x&1\\y&x\end{array}\right)\right|=\left|y+2x^2\right|=y+2x^2\ \
\mbox{in the first quadrant.} $$ Therefore
\begin{eqnarray*}
I&=&\int_1^2\, dv\int_0^1(5u^4+6v)\,
du=\int_1^2\left[u^5+6uv\right]_{u=0}^1\, dv\\ &=&\int_1^2(1+6v)\,
dv =\left[v+3v^2\right]_1^2 =10.
\end{eqnarray*}

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