- Thread starter
- Admin
- #1

I have posted a link there to this topic so the OP can see my work.The volume of the solid generated by revolving the region bounded by the graph y=x^2,y^2=8x,about the y-axis?

- Thread starter MarkFL
- Start date

- Thread starter
- Admin
- #1

I have posted a link there to this topic so the OP can see my work.The volume of the solid generated by revolving the region bounded by the graph y=x^2,y^2=8x,about the y-axis?

- Thread starter
- Admin
- #2

First, let's determine the points of intersection for the two curves. Substituting for $y$ as given in the first equation into the second, we obtain:

\(\displaystyle \left(x^2 \right)^2=8x\)

\(\displaystyle x^4-8x=0\)

\(\displaystyle x\left(x^3-2^3 \right)=0\)

\(\displaystyle x(x-2)\left(x^2+2x+2^2 \right)=0\)

The quadratic factor has complex roots, hence we find:

\(\displaystyle x=0,\,2\)

Thus, the two points of intersection are:

\(\displaystyle (0,0),\,(2,4)\)

To use the washer method, we find the volume of an arbitrary washer is:

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

where:

\(\displaystyle R^2=y\)

\(\displaystyle r^2=\frac{y^4}{64}\)

Hence:

\(\displaystyle dV=\pi\left(y-\frac{y^4}{64} \right)\,dy\)

Adding the volume elements by integrating, we find:

\(\displaystyle V=\pi\int_0^4 y-\frac{y^4}{64}\,dy\)

Applying the FTOC, we obtain:

\(\displaystyle V=\pi\left[\frac{y^2}{2}-\frac{y^5}{320} \right]_0^4=\frac{24\pi}{5}\)

Using the shell method, we find the volume of an arbitrary shell is:

\(\displaystyle dV=2\pi rh\,dx\)

where:

\(\displaystyle r=x\)

\(\displaystyle h=\sqrt{8x}-x^2\)

Hence:

\(\displaystyle dV=2\pi x\left(\sqrt{8x}-x^2 \right)\,dx=2\pi\left(2\sqrt{2}x^{\frac{3}{2}}-x^3 \right)\,dx\)

Summing the volume elements by integrating, we find:

\(\displaystyle V=2\pi\int_0^2 2\sqrt{2}x^{\frac{3}{2}}-x^3\,dx\)

Applying the FTOC, we find:

\(\displaystyle V=2\pi\left[\frac{4\sqrt{2}}{5}x^{\frac{5}{2}}-\frac{x^4}{4} \right]_0^2=2\pi\left(\frac{32}{5}-4 \right)=\frac{24\pi}{5}\)