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$y = x^3, y = 0, x = 1;$ about $x = 2$

I understand that this is a cross sectional graph and I know how to do it with two curves, but the $y = 0$ throws me off. I am left with one curve so I assume the integral would be:

$\int \pi(2-x^3)^2dx$

but I'm not getting the correct answer. Maybe I calculated wrong. Could someone assist me with this?

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2 Answers 2

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Well, first you need to define the area that is to be rotated. You do need all three curves to do that. It's a good idea to sketch them very roughly...

The area is bounded on the bottom by $y=0$, on the right by the vertical line $x=1$. The third "side" is the curve $y=x^3$. So, the three "corners" of the area are $(0,0)$, $(1,0)$, and $(1,1)$.

Next, what is the axis of rotation? Easy, it's the vertical line $x=2$.

Now, you need to decide whether to use shells or discs; let's use discs. Or rather washers, The tricky bit is that the axis of rotation is distinct from the area being rotated. The final volume is basically a volcano cone minus the crater...

The outer radius of the washer is $2-x$, the inner radius is $2-1=1$, the thickness is $dy$, so the volume of the washer is $$pi((2-x)^2-1^2)dy$$ Replace x with $y^{(1/3)}$ and add up all the washers from $y=0$ to $y=1$

General question: is this too much help? I'm new here...

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The region is not really well described, but it it likely that you are to rotate the finite region below $y=x^3$, above $y=0$, and to the left of $x=1$ about the line $x=2$.

If that is the case, the method usually called the method of cylindrical shells is a natural approach. Take a thin vertical slice of width going from $x$ to $x+dx$. this slice has height about $x^3$. When it is roated about $x=2$, we get a shell of radius $2-x$, with volume $2\pi(2-x)x^3\,dx$. "Add up" (integrate) from $x=0$ to $x=1$. We get $$\int_0^1 2\pi(2-x)x^3\,dx.$$ This is probably $\frac{3\pi}{5}$.

We could alternately use the method of slicing (cross-sections). In that case we will be integrating with respect to $y$. At height $y$, the cross-section is a circle of radius $2-x$, with a hole of radius $1$ drilled out. We have $x=y^{1/3}$, so our volume is $$\int_{y=0}^1 \pi\left((2-y^{1/3})^2-1^2\right)\,dy.$$ The algebra is somewhat more unpleasant, but again we get $\frac{3\pi}{5}$.

Remark: The formula proposed in the post is sort of related to rotating about the line $y=2$. For that, we would want $\int_0^1 \pi\left(2^2-(2-x^3)^2\right)\,dx$.

It is usually not very hard to solve these rotation problems, if we draw a picture and each time go back to basics. Using a remembered formula is, for me at least, far less reliable.

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