I'm reviewing for a final exam and found a great question that I would like some assistance with!

Write two integrals representing the volume of the solid obtained by rotating the region ecnlsoed by the $$y-axis$$ $$y=\sqrt{x}$$ $$y=2-x^3$$ about the line $$y=2$$ One integral should be with respect to x while the other should be with respect to y.

So when I did this problem I was rotating the solid to the right using the shell method, but I think I was doing the problem totally wrong. It should be rotating upwards and I shouldn't need to use the shell method, but I'm not sure how to do it without the shell method. Any explanations would be fantastic! Thanks!


Draw a careful diagram, and identify visually the region we are rotating. The analyses below won't make much sense without the picture.

Note that the two curves $y=\sqrt{x}$ and $y=2-x^3$ meet at $x=1$. (There is no pleasant "algebraic" way to solve the equation $\sqrt{x}=2-x^3$, but we can see by inspection that $x=1$ works. And it is clear that there is only one real solution, since $\sqrt{x}$ is increasing and $2-x^3$ is decreasing.)

Slicing: We take cross-sections perpendicular to the $x$-axis. We get a "washer" (a disk with a hole in it), The outer radius of the washer is $2-\sqrt{x}$, and the inner radius is $2-(2-x^3)$, which is $x^3$. Thus the area of cross-section is $\pi\left((2-\sqrt{x})^2-(x^3)^2\right)$. It follows that the volume is $$\int_{x=0}^1 \pi\left((2-\sqrt{x})^2-(x^3)^2\right)\,dx.$$

Shells: Take a thin slice at height $y$, going from $y$ to $y+dy$, and rotate it about the line $y=2$. We get a cylindrical shell. The radius of the shell is always $2-y$. But the height of the cylinder varies, depending on whether $0\le y\le 1$ or $1\le y\le 2$.

If $0\le y\le 1$, the height of the shell is $x$, which is $y^2$. Integrating from $y=0$ to $y=1$, we get a contribution of $$\int_{y=0}^1 2\pi(2-y)y^2\,dy\tag{1}$$ to the volume. The analysis is much the same from $y=1$ to $y=2$, but here $x=(2-y)^{1/3}$. That gives a contribution of $$\int_{y=1}^2 2\pi(2-y)(2-y)^{1/3}\,dy\tag{2}$$ to the volume. finally, add integrals (1) and (2).

Remark: In your picture, you will have drawn the horizontal line $y=2$. The region of interest is below the line, so yes, we are rotating upwards.

Doing it by Shells is sort of necessary. Slicing is easier, but the question asked for two methods.

  • $\begingroup$ Thanks for the detailed explanation, I fully understand now! $\endgroup$ – user133707 May 7 '14 at 3:16
  • $\begingroup$ You are welcome. Usually everything becomes clear once we have a clear view of the geometry. $\endgroup$ – André Nicolas May 7 '14 at 13:06

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