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{\bf Question}

Sketch the region enclosed by the given curves and find the volume
of the solid generated when it is revolved about the $x$-axis:

$xy=4$, $x+y=5$.

{\bf Answer}

At which points does the line $x+y=5$ intersect the curve $xy=4$?
Writing $y=5-x$ we have $xy=x(5-x)=4$ and so $5x-x^2=4$ or
$x^2-5x+4=0$.  This factorises to $(x-1)(x-4)=0$ and hence $x=1$
or $x=4$.  Using $y=5-x$ we have the points $(x,y)=(1,4)$ and
$(x,y)=(4,1)$.

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Use washer method: \begin{eqnarray*} & & \int_{x=1}^{x=4} \left
\{\pi(5-x)^2-\pi\left(\frac{4}{x}\right)^2 \right\} \,dx\\ & = &
\pi \int_{x=1}^{x=4} \left \{(5-x)^2 -\frac{16}{x^2}\right\}
\,dx\\ & = & \pi
\left[-\frac{(5-x)^3}{3}+\frac{16}{x}\right]_{x=1}^{x=4}\\ & = &
\pi
\left\{\left(-\frac{1}{3}+4\right)-\left(-\frac{4^3}{3}+16\right)
\right\} \\ & = & 9\pi. \end{eqnarray*}

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