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QUESTION An area $A$ is enclosed between the curve
$y=x^\frac{1}{3}$, the $x$-axis and the lines $x=0$ and $x=8$.

\begin{description}

\item[(i)]
Find the magnitude of $A$.

\item[(ii)]
Calculate the coordinates of the centroid of $A$.

\item[(iii)]
Find the volume generated when $A$ is rotated about the $y$-axis.

\end{description}



ANSWER

DIAGRAM

\begin{description}

\item[(i)]
$\ds
A=\int_0^8y\,dx=\int_0^8x^{\frac{1}{3}}\,dx=\left[\frac{x^\frac{4}{3}}{\frac{4}{3}}\right]_0^8
=\frac{3}{4}(8^\frac{4}{3}-0)=\frac{3}{4}(2^4)=12$

\item[(ii)]
The coordinates of the centroid are $(\overline{x},\overline{y})$.

$\ds A\overline{x}=\int_0^8xy\,dx=\int_0^8x^\frac{4}{3}\,dx=
\left[\frac{x^\frac{7}{3}}{\frac{7}{3}}\right]_0^8
=\frac{3}{7}(8^\frac{7}{3})=\frac{3}{7}(2^7)$
\\therefore $\ds\overline{x}=\frac{3(128)}{7(12)}=\frac{32}{7}\\
A\overline{y}=\int_0^8\frac{1}{2}y^2\,dx=
\int_0^8\frac{1}{2}x^\frac{2}{3}\,dx= \frac{1}{2}\left[\frac{
x^\frac{5}{3}}{\frac{5}{3}}\right]_0^8$\\ i.e. $\ds
A\overline{y}=\frac{3}{10}(8^\frac{5}{3})=\frac{3(32)}{10}$\\
Therefore $\ds\overline{y}=\frac{3(32)}{10(12)}=\frac{4}{5}$

\item[(iii)]
When $B$ is rotated about the $y$-axis the resulting volume is
given by $\ds\int_0^2\pi x^2\,dy=\int_0^2\pi
y^6\,dy=\pi\left[\frac{y^7}{7}\right]_0^2=\frac{128\pi}{7}$\\
Hence the volume when $A$ rotated is that of a cylinder $-$ the
above volume, i.e. \\ $\ds (\pi8^2)2-\frac{128\pi}{7}
=128\pi-\frac{128\pi}{7}= 128\pi\times\frac{6}{7}
=\frac{768\pi}{7}(=344.68)$

\end{description}



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