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Consider the region bounded by $y^2=x^3$, the $x$ axis and $x=4$. I want to calculate the volume of the solid of revolution of the region around $y=8$ and I can use the washer method, as the outer radius is $8-0=8$ and the inner radius is $8-x^\frac{3}{2}$. So let's calculate:

$$V_1=\pi\int_0^4 [8^2-(8-x^\frac{3}{2})^2]dx=\frac{704}{5}\pi u^3.$$ but then I'm struggling and confused as I get this result from Wolfram Alpha:

$$V_2=∫_0^4 2 π (4-x) (8-x^\frac{3}{2})dx=\frac{3456}{35}\pi u^3.$$

Why do I get a different result?

And I have another related question. I was wandering if I could get a nice plot (visualization) of the same solid of revolution. As Wolfram Alpha is giving a different result, I can't trust the plot given by the computation linked above and also showed below.

Wolfram Alpha all credits

Thank you!

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The integral $V_2$ is for the volume when we rotate the region below $y=8$ and above $y=x^{3/2}$ about the vertical line $x=4$. It uses the method of cylindrical shells. So note that (i) the region is different and (ii) the axis of rotation is different.

You noted in a comment that when we rotate the region between $y=x^{3/2}$ and $x=0$ about $x=4$, we get volume $\frac{1024}{35}\pi$. When we add the volume cpmputed by Alpha, the sum simplifies to $128\pi$, the volume of a cylinder radius $4$ and base $8$.

The integral $V_1$ computes the volume when we rotate the region below $y=x^{3/2}$ and above the $x$-axis about the line $y=8$. This is the region and axis of rotation specified in the problem.

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  • $\begingroup$ The volume of the region when we rotate it about $x=4$ should be $\frac{1024}{35}\pi u^3$ as $V=∫_0^8 (4-y^\frac{2}{3})^2dy=\frac{1024}{35}\pi u^3.$. $\endgroup$ – Lorenzo Comoglio Jun 5 '16 at 0:24
  • $\begingroup$ @LorenzoComoglio: Yes, the region Alpha is rotating is different, as is the axis of rotation. $\endgroup$ – André Nicolas Jun 5 '16 at 0:29
  • $\begingroup$ Can you please show me a plot of the solid which volume is $V_1$? $\endgroup$ – Lorenzo Comoglio Jun 5 '16 at 0:33
  • $\begingroup$ I still don't get why Wolfram Alpha is giving the result taking another region, I mean the region above the curve, as you stated. The input is correct... $\endgroup$ – Lorenzo Comoglio Jun 5 '16 at 0:36
  • $\begingroup$ @LorenzoComoglio: I do not have familiarity with 3-D graphing software. But you seem to have visualized the solid described by the problem correctly, since you got the right integral. $\endgroup$ – André Nicolas Jun 5 '16 at 0:40

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