Hi,
I have done the question on the picture attached
I don't believe it could be the correct answer, can someone help me please
Thanks
$\displaystyle V = \pi \int_0^3 (x^2+4x)^2 \, dx
$
$\displaystyle V = \pi \int_0^3 x^4 + 8x^3 + 16x^2 \, dx
$
$\displaystyle V = \pi \left[\frac{x^5}{5} + 2x^4 + \frac{16x^3}{3}\right]_0^3
$
$\displaystyle V = \pi \left[\frac{3^5}{5} + 2 \cdot 3^4 + \frac{16 \cdot 3^3}{3}\right]
$
$\displaystyle V = \frac{1773\pi}{5} \approx 1114$
Hi wolfhound.
It always helps to understand how the situation looks on a diagram.
The volume of revolution is the sum of the areas of a series of "wafer-thin" discs (an "infinite" sum of them, evaluated as their widths "deta-x" goes to zero).
The radius of the discs in this case is f(x) because the curve is being rotated around the x-axis.
the area of a disc is $\displaystyle {\pi}r^2={\pi}[f(x)]^2$
All of these disc areas are integrated from x=0 to x=3.
This gives the volume of revolution.
You need to get used to how to do this notationally,
as your written work needs to be redone.
$\displaystyle Disc\ area={\pi}\left(x^2+4x\right)^2={\pi}\left(x^4+8x^ 3+16x^2\right)$
Therefore, the volume of revolution is
$\displaystyle \int_{x=0}^{x=3}{\pi}\left(x^4+8x^3+16x^2\right)dx$
$\displaystyle ={\pi}\left[\frac{x^5}{5}+\frac{8x^4}{4}+\frac{16x^3}{3}\right]$ from x=0 to 3
I've shown the disc from an angle (hence it looks like an ellipse!)