Finding the volume of a cylinder

[Geekchick]

Homework Statement

x²+y²+16

Homework Equations

\pi\int^{b}_{a}{R(x)²}dx

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

 

[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.

 

[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.