Find the volume of the solid obtained by rotating it about the x-axis?

[randoreds]

Hey guys, I know it late its a little past one here. But I'm doing an assignment due tomorrow at I've been stuck on the last question for at least an hour.

Find the volume of the solid obtained by rotating the region bounded the curves

Y=absolute value of x. and y = square root of ( 8-x^2) , about the x-axis.

its going to look like a hemisphere with like a cone cut out. I just have no idea how to remove the cone!

I promise if you help, the world will bless you with karma.

Its funny that last poster is doing the same type of problem, but we're not in the same class because I wasn't assigned that problem! I think the calculus world just decided today all calculus classes find volumes!

Thanks!

 

[aralbrec]

its going to look like a hemisphere with like a cone cut out. I just have no idea how to remove the cone!

Find the volume of the curve, subtract the volume of the cone. You can keep it in one integral by noting that A(x) is the area of the curve minus the area of the cone.

karma++ and time for bed :)

 

[bossman27]

To make it easier, I might do this by only thinking about half of the volume on one side of the y-axis first, then just multiplying by two.

But either way, draw your graph, find the intersection point(s). Do you know how to find the volume by rotation using the area enclosed by one function?

If you do, the case of an area enclosed by two functions isn't too bad, especially when they don't cross over the interval we're concerned with. All you need to know is:

V = V(outer radius) - V(inner radius)

See how far you can get with that first.