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"A region is enclosed by the curves $y=x$ and $y=x^{2}$. Using the Washer method, find the volume by rotating this region about the line $y=2$."

This is a homework problem of mine and frankly, I don't even know where to begin. Is there someone here who could possible give me an intuitive explanation of how to go about solving this?

Will this object be rotating about the $y$-axis at $y=2$? So we begin our integral at 2 and end it at infinity?

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The line $y=2$ is not parallel to the $y$-axis. – Muphrid Dec 12 '12 at 6:18
up vote 2 down vote accepted

It is helpful to first draw a picture: enter image description here

Notice that $x = x^2$ implies that $x^2 - x = 0$ so that $x = 0$ or $x = 1$. So we are rotating the area around the line $y = 2$. We think of the volume we are seeking as a difference of two volumes. First, we rotate the outermost function ($\color{blue}{y = x^2}$) around the line $y = 2$. Then, we rotate the innermost function ($\color{green}{y = x}$) around the line $y = 2$. Taking the difference will give the volume we are seeking:

$$ V = \pi \int_0^1 (2 - x^2)^2 dx - \pi \int_0^1 (2 - x)^2 dx = \pi\int_0^1 ((2 - x^2)^2 - (2 - x)^2)dx $$

Again, to give you a better idea of what we are doing, we start by partitioning the area between the curves with rectangles. Next, imagine we pick up each rectangle and rotate it around the line $y = 2$. When we do this with each rectangle, we get a solid that looks like a washer (hence the name). Finally, the volume of a disk of width $\Delta x$ and height $y = f(x)$ is $\pi f(x)^2 \Delta x$. Adding these up give:

$$ V \approx \sum_{k=1}^n \pi f(x_k)^2 \Delta x \to \pi \int_a^b f(x)^2 dx \qquad (\text{as }n \to \infty) $$

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That helped tremendously, thank you so much. – Kyle H Dec 12 '12 at 6:45
@KyleH: There was a typo in my original answer. Please see the corrected version. – JavaMan Dec 12 '12 at 6:45
Ahh ok. Again, Thank you for explaining this to me. – Kyle H Dec 12 '12 at 23:14
Not a problem. I'm glad it helped. – JavaMan Dec 12 '12 at 23:20

This post might be quite helpful:(and I do not think I can explain any better)

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