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 $\displaystyle x=3$ to $\displaystyle x=0$
Attempt: If $\displaystyle x=3$, the height of the generated volume is also 3.
For $\displaystyle x=3$, $\displaystyle y=2x\rightarrow y=2(3)\rightarrow y=6$. Thus, the height of the volume generated is 6.
Volume: $\displaystyle \pi\int^3_0y^2\,dx=\pi\int^3_0\frac{r^2(2x)^2}{h^2 }=\pi\int^3_0\frac{r^2(4x^2)}{h^2}=\pi[\frac{r^2(4x^3}{3(h^2)}]^3_0=\pi[\frac{6^2(4(3)^3)}{3(3)^2}]-\pi[\frac{6^2(4(0)^3)}{3(3)^2}]=\pi\frac{3888}{27}-0=144\pi$
Ans.: (From text book): $\displaystyle 36\pi$
