Volumes of Revolution

**GrigOrig99** · December 10th 2012, 11:49 AM

Having a bit of trouble with this one. Can anyone help me out?

Many thanks.

Q. The given diagram (see attachment) shows the line [tex]y=2x[tex]. Find the volume generated when this line is rotated about the x-axis from $x=3$ to $x=0$

Attempt: If $x=3$ , the height of the generated volume is also 3.
For $x=3$ , $y=2x\rightarrow y=2(3)\rightarrow y=6$ . Thus, the height of the volume generated is 6.
Volume: $\pi\int^3_0y^2\,dx=\pi\int^3_0\frac{r^2(2x)^2}{h^2 }=\pi\int^3_0\frac{r^2(4x^2)}{h^2}=\pi[\frac{r^2(4x^3}{3(h^2)}]^3_0=\pi[\frac{6^2(4(3)^3)}{3(3)^2}]-\pi[\frac{6^2(4(0)^3)}{3(3)^2}]=\pi\frac{3888}{27}-0=144\pi$

Ans.: (From text book): $36\pi$

**hairymclairy** · December 10th 2012, 12:41 PM

Hey,

Have a think about the shape that is made when the line $y=2x$ is rotated through $2\pi$ around the x-axis. There is a formula for calculating the volume of such a shape (hint: $area=\frac{\pi r^2h}{3}$ , where h is height and r is radius).

**GrigOrig99** · December 10th 2012, 12:42 PM

Ok, I think I see where I went wrong. Thank you.

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