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QUESTION


\begin{description}

\item[(a)]
Evaluate the double integral
$\begin{displaystyle}\int_0^2\!\int_{y^2}^2y(2x-y^2/4)\,dxdy\end{displaystyle}$.

\item[(b)]
Given the integral
$\begin{displaystyle}\int_0^4\!\int_{\sqrt{y}}^2y\sin
x^5\,dxdy\end{displaystyle}$, sketch the region of integration.
Reverse the order of integration and evaluate the resulting
integral.

\item[(c)]
Sketch the plane region $R$ defined by the inequalities, computing
the intercepts of the defining curves with the $x$ and $y$ axes
and with each other:

$$x\leq\sqrt{y+6},\ x\geq\sqrt{2y},\ y\geq0.$$

By choosing an appropriate order of integration show how to
express an integral of the form $\int\!\!\!\int_Rf(x,y)\,dA$ as a
double integral, computing the limits.

\end{description}



ANSWER


\begin{description}

\item[(a)]

\begin{eqnarray*}
\int_0^2\!\int_{\sqrt{y}}^2y(2x-y^2/4)\,dxdy&=&\int_0^2\left[x^2y-x\frac{y^3}{4}\right]_{y^2}^2\\
&=&\int_0^2\left[4y-\frac{2y^3}{4}-y^5+\frac{y^5}{4}\right]\,dy\\
&=&\left[4\frac{y^2}{2}-\frac{y^4}{8}-\frac{y^6}{6}+\frac{y^6}{24}\right]_0^20\\
&=&-2
\end{eqnarray*}

\item[(b)]

DIAGRAM

\begin{eqnarray*}
\int_0^2\!\int_0^{x^2}y\sin x^5\,dydx\\
&=&\int_0^2\left[\frac{y^2}{2}\sin
x^5\right]_0^{x^2}=\int_0^2\left[\frac{x^4}{2}\sin x^5\right]\,dx
\end{eqnarray*}

put $u=x^5$ so $\frac{du}{dx}=5x^4$ to get

\begin{eqnarray*}
\int_0^2\frac{1}{10}\sin u\,du&=&\left[-\frac{1}{10}\cos
u\right]^{x=2}_{x=0}\\ &=&-\frac{1}{10}\cos32+\frac{1}{10}
\end{eqnarray*}

\item[(c)]

DIAGRAM

$$\int\!\!\!\int_Rf(x,y)dA=\int_0^6\int_{\sqrt{2y}}^{\sqrt{y+6}}f(x,y)\,dxdy$$

\end{description}




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