Finding the volume of a cylinder

1. Feb 1, 2009

Geekchick

1. The problem statement, all variables and given/known data

x2+y2+16

2. Relevant equations

$$\pi$$$$\int^{b}_{a}$${R(x)2}dx

3. The attempt at a solution

I just need to know if i set this up right.

When I solve for y and graph it I get a semi circle that goes from -4 to 4.

$$\pi\int^{0}_{-4}$${($$\sqrt{(16-x^{2})}$$)$$^{2}$$}dx + $$\pi\int^{4}_{0}$${($$\sqrt{(16-x^{2})}$$)$$^{2}$$}dx

I get 85.4$$\pi$$

2. Feb 1, 2009

Gregg

You have given an expression but i'll assume you mean the equation for:

$x^2 + y^2 - 16 = 0$

which is a circle whose centre is at 0,0 and has a radius of 4 units?

The area of such a cylinder would be

$16\pi l$

where l is the length of the cylinder.

I don't think you need to use $\int_a^b y^2 dx$ because you know the radius and it's not a volume of revolution.

If you want to find the sphere when the shape is rotated about the y axis pi radians you can just use the formula

$\frac{4}{3} \pi r^3$

$\frac{256}{3} \pi$

$85.3 \pi$

So no need for integration, but you got the right answer.

3. Feb 1, 2009

Dick

It looks to me like you are trying to find the volume of a SPHERE by rotating a semicircle. So yes, the answer is correct. But no need to round it off or to break the integral into two parts. Like Gregg said it's pi*256/3. And the equation is x^2+y^2=16. This has nothing to do with cylinders.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook