18 Find the volume of the solid generated by revolving the region about y-axis.

[karush]

$\textsf{Find the volume of the solid generated by revolving the region
about y-axis.Given the boundares of }\\$
\begin{align*} \displaystyle
y&=4x-x^2\\
y&=x\\
\end{align*}

Ok, I presume since this is rotated around the y-axis that we have to rewrite the equations in terms of y

$y=-(x^2-4x)$
$4-y=(x-2)^2$
$\sqrt{4-y}=x-2$
$\sqrt{4-y}+2=x$

and

$x=y$

But the graph is not complete and is this not the same volume if it were rotated around the x-axis.

I thought this was going to be very easy, but going around the y-axis was confusing.

 

[karush]

View attachment 7491

pic of graph

 

[MarkFL]

I would use the shell method, so you don't have to use two integrals...what is the radii of the shells...what is the height?

 

[karush]

\begin{align*}
\displaystyle
\int_0^3 2\pi x (3x-x^2)dx& = 2\pi \int_0^3 (3x^2-x^3)dx\\
& = 2\pi \left[x^3-\dfrac{x^4}{4}\right]_0^3 \\
&= 2\pi\left(27-\dfrac{81}{4}\right)\\
&= \dfrac{27\pi}{2}
\end{align*}

I followed an example in book!
hope its right

 

[MarkFL]

\begin{align*}
\displaystyle
\int_0^3 2\pi x (3x-x^2)dx& = 2\pi \int_0^3 (3x^2-x^3)dx\\
& = 2\pi \left[x^3-\dfrac{x^4}{4}\right]_0^3 \\
&= 2\pi\left(27-\dfrac{81}{4}\right)\\
&= \dfrac{27\pi}{2}
\end{align*}

I followed an example in book!
hope its right

Let's check by using the washer method:

\(\displaystyle dV=\pi\left(R^2-r^2\right)\,dy\)

On $[0,3)$:

\(\displaystyle R=y,\,r=2-\sqrt{4-y}\)

On $[3,4]$:

\(\displaystyle R=2+\sqrt{4-y},\,r=2-\sqrt{4-y}\)

Hence:

\(\displaystyle V=\pi\left(\int_0^3 y^2-\left(2-\sqrt{4-y}\right)^2\,dy+\int_3^4 \left(2+\sqrt{4-y}\right)^2-\left(2-\sqrt{4-y}\right)^2\,dy\right)\)

\(\displaystyle V=\pi\left(\int_0^3 y^2-\left(8-y-4\sqrt{4-y}\right)\,dy+\int_3^4 \left(4\right)\left(2\sqrt{4-y}\right)\,dy\right)\)

\(\displaystyle V=\pi\left(\int_0^3 y^2+y+4\sqrt{4-y}-8\,dy+8\int_3^4 \sqrt{4-y}\,dy\right)\)

\(\displaystyle V=\pi\left(\left[\frac{y^3}{3}+\frac{y^2}{2}-8y\right]_0^3+\frac{8}{3}\left(\left[u^{\frac{3}{2}}\right]_1^4+2\left[u^{\frac{3}{2}}\right]_0^1\right)\right)\)

\(\displaystyle V=\pi\left(9+\frac{9}{2}-24+\frac{8}{3}\left((8-1)+2\right)\right)=\frac{27}{2}\pi\quad\checkmark\)