# Math Help - Volume of a solid

1. ## Volume of a solid

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?

2. Originally Posted by jigger1
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 $8-0.5x^2$. The form for cylindrical shells is $\int_{a}^{b} 2\pi{r}{h(r)}dr$. In this case, you will integrate $\int_{0}^{4} 2\pi{x}{(8-0.5x^2)}dx$.

3. $y = 8 - \frac{1}{2}x^{2}$ intersects the x-axis at x = 4.

So by method of cylindrical shells, we have:
$V = \int_{0}^{4} A(x) dx$ where A(x) is the surface area of the cylinder
$V = \int_{0}^{4} 2\pi r h$

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:
$V = 2\pi \int_{0}^{4} x \left(8 - \frac{1}{2}x^{2}\right)dx$

Edit: Ok too slow xD

4. Thanks icemanfan. That gets me started.

5. To o_O: But your explanation is better than mine

6. Can any one assist/explain how to solve from here?

7. Originally Posted by jigger1
Can any one assist/explain how to solve from here?
Evauate the integral

$V = 2\pi \int_{0}^{4} x \left(8 - \frac{1}{2}x^{2}\right)dx$

$V = 2\pi \int_{0}^{4} \left(8x - \frac{1}{2}x^{3}\right)dx=2\pi\left( 4x^2-\frac{1}{8}x^4\right)|_0^{4}$

$2\pi(64-32)=64\pi$

8. We could also use washers. Solve the given for x and we get

$x=\sqrt{16-2y}$

${\pi}\int_{0}^{8}(16-2y)dy$