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

MarkFL

Staff member
Here is the question:

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

Staff member
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 $T$: 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:

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