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I'm stuck on a problem from my Calc 2 class and I would appreciate some help with it.

My object is bounded by $y=\sqrt{x}$, $y=0$, and $x=3$ and rotated about the y-axis. Here is what I have setup:

$$v=\pi\int_0^{\sqrt{3}}{y^2}^2 dy$$


The book says it is $$v=\frac{\pi36\sqrt{3}}{5}$$

Thanks in advance

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2 Answers 2

up vote 5 down vote accepted

I intended to make a comment but I made a typo and it came too long, so...

You're integrating over the wrong area. Do a drawing, you want to revolve around the $y$-axis the area between the graph of $x= y^2$ and $x=3$, and you're doing the area between $x=0$ and $x=y^2$ !

The expression representing the volume of revolution should be

$$\pi\int\limits_0^{\sqrt 3}(9-y^4)dy=9\sqrt 3\,\pi-\pi\int\limits_0^{\sqrt 3}y^4\,dx=9\sqrt 3\,\pi-\left.\frac{\pi}{5}y^5\right|_0^{\sqrt 3}=9\sqrt 3\,\pi-\frac{3^{5/2}\pi}{5}=\frac{36\sqrt 3\pi}{5}$$

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Thank you so much, I wouldn't have done that if it was x values :) –  David Feb 1 '13 at 22:56
I know, it may be confusing. Drawing the graph of the gang of functions involved and reading carefully what area the questions asks to be revolved helps a lot. –  DonAntonio Feb 2 '13 at 0:30
+1 This answer needs additional stars. ;) –  Babak S. Feb 2 '13 at 11:41

In this case, it may be easier to use the method that is often called the Method oc Cylindrical Shells. Take a thin vertical strip between $x$ and $x+dx$. Its area is roughly $\sqrt{x}\,dx$. When rotated about the $y$-axis, it sweeps out a shell of volume approximately $(2\pi x)x^{1/2}\,dx$. "Add up" (integrate) from $0$ to $3$. So we want $$\int_0^3 2\pi x^{3/2}\,dx.$$ The integral is $(2\pi)\frac{2}{5}3^{5/2}$.

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Yes but I was required to use the disk method for this section +1 though –  David Feb 2 '13 at 3:37
@David: "Holes" like the one we get by disk/washer method sometimes cause students trouble. That's why I pointed at the shells method, which sidesteps that issue. –  André Nicolas Feb 2 '13 at 3:41

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