A solid is formed when the area bounded by y=8-0.5x^2, x=0 and y=0 is rotated about the y axis. Determine the volume of the solid.
How do I go about solving this?
I would use cylindrical shells to solve this problem. The height of each shell will be $\displaystyle 8-0.5x^2$. The form for cylindrical shells is $\displaystyle \int_{a}^{b} 2\pi{r}{h(r)}dr$. In this case, you will integrate $\displaystyle \int_{0}^{4} 2\pi{x}{(8-0.5x^2)}dx$.
$\displaystyle y = 8 - \frac{1}{2}x^{2}$ intersects the x-axis at x = 4.
So by method of cylindrical shells, we have:
$\displaystyle V = \int_{0}^{4} A(x) dx$ where A(x) is the surface area of the cylinder
$\displaystyle V = \int_{0}^{4} 2\pi r h$
If you make a quick sketch of the graph, you'll see it's an inverted parabola. If you imagine drawing one of the cylinders from the y-axis, you'll see that the radius is the horizontal distance from the y-axis to the curve which is just simply x. The height of the cylinder is the distance from the x-axis to the curve (which is simply the curve it self).
So:
$\displaystyle V = 2\pi \int_{0}^{4} x \left(8 - \frac{1}{2}x^{2}\right)dx$
Edit: Ok too slow xD