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

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In summary, the volume of the solid of revolution formed when the region T, enclosed between the graphs of y=x^2 and y=x^3, is revolved about the y-axis can be found using both the washer and shell methods. Both methods result in a volume of \pi/10.
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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.
 
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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:

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

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

1. How do the washer and shell methods differ?

The washer method involves finding the volume of a solid of revolution by slicing it into thin, hollow disks. The shell method involves finding the volume by slicing the solid into thin, hollow cylinders.

2. When should I use the washer method vs the shell method?

The washer method is typically used when the axis of revolution is perpendicular to the axis of integration, while the shell method is often used when the axis of revolution is parallel to the axis of integration.

3. What is the formula for the washer method?

The formula for the washer method is V = π∫(outer radius)^2 - (inner radius)^2 dx.

4. Can the washer and shell methods be used for any shape?

Yes, the washer and shell methods can be used for any shape that can be rotated around an axis.

5. Are there any limitations to using the washer and shell methods?

The main limitation is that the shape must be revolved around an axis. Additionally, the method may become more complicated for shapes with varying cross-sections or for certain types of integrals.

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