Martin's question at Yahoo Answers regarding using the washer and shell methods

[MarkFL]

Here is the question:

Calculus volume help please?

Let T be the region enclosed between the graphs of y=x^2 and y=x^3

A. Use the washer method to find the volume of the solid of revolution formed when T is revolved about the Y-axis.

B. Use the shell method to compute the volume of the solid of revolution described in Part A.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello martin,

First, let's determine the coordinates of the intersections of the given curves. Equating them we find:

\(\displaystyle x^3=x^2\)

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

\(\displaystyle x^2(x-1)=0\)

Thus, we find the two points are $(0,0)$ and $(1,1)$. Here is a diagram of the region $T$:

View attachment 1933

a) Use the washer method to find the volume of the solid of revolution formed when $T$ is revolved about the $y$-axis.

The volume of an arbitrary washer is:

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

where:

\(\displaystyle R=y^{\frac{1}{3}}\)

\(\displaystyle r=y^{\frac{1}{2}}\)

And so we have:

\(\displaystyle dV=\pi\left(y^{\frac{2}{3}}-y \right)\,dy\)

Summing up all the washers, there results:

\(\displaystyle V=\pi\int_0^1 y^{\frac{2}{3}}-y\,dy\)

Applying the FTOC, we get:

\(\displaystyle V=\pi\left[\frac{3}{5}y^{\frac{5}{3}}-\frac{1}{2}y^2 \right]_0^1=\pi\left(\frac{3}{5}-\frac{1}{2} \right)=\frac{\pi}{10}\)

b) Use the shell method to compute the volume of the solid of revolution described in part a).

The volume of an arbitrary shell is:

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

where:

\(\displaystyle r=x\)

\(\displaystyle h=x^2-x^3\)

And so we have:

\(\displaystyle dV=2\pi\left(x^3-x^4 \right)\,dx\)

Summing up all of the shells, there results:

\(\displaystyle V=2\pi\int_0^1 x^3-x^4\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle V=2\pi\left[\frac{1}{4}x^3-\frac{1}{5}x^5 \right]_0^1=2\pi\left(\frac{1}{4}-\frac{1}{5} \right)=\frac{\pi}{10}\)