Volume of a solid between 2 functions revolved about the x-axis

[sakau2007]

Homework Statement

Find the volume of a solid bounded by the functions y=1-x² and y=0 revolved around the x-axis

Homework Equations

(don't know how to show integrals)
A=Pi * the integral from x_a to x_b of [f(x)]²-[g(x)]²

The Attempt at a Solution

First, to find the lower and upper x-bounds, set the functions equal to one another to get their points of intersection. You get x = 1 and -1.

To find which function is f(x) (the greater values), pick a test point between -1 and 1. I used 0 and found 1-x² was greater along this interval.

So, V=Pi* the integral from -1 to 1 of (1-x²)² dx.
After FOIL I get 1-2x^2+x4. The integral of this polynomial is x-(2/3)x³+(1/5)x⁵. I then evaluated this by plugging in 1 and subtracting when I plug in -1.

I get 16pi/15. Unsure if this is correct. Would like if someone could check my work.

 

[Dick]

Homework Statement

Find the volume of a solid bounded by the functions y=1-x² and y=0 revolved around the x-axis

Homework Equations

(don't know how to show integrals)
A=Pi * the integral from x_a to x_b of [f(x)]²-[g(x)]²

The Attempt at a Solution

First, to find the lower and upper x-bounds, set the functions equal to one another to get their points of intersection. You get x = 1 and -1.

To find which function is f(x) (the greater values), pick a test point between -1 and 1. I used 0 and found 1-x² was greater along this interval.

So, V=Pi* the integral from -1 to 1 of (1-x²)² dx.
After FOIL I get 1-2x^2+x4. The integral of this polynomial is x-(2/3)x³+(1/5)x⁵. I then evaluated this by plugging in 1 and subtracting when I plug in -1.

I get 16pi/15. Unsure if this is correct. Would like if someone could check my work.

Looks fine to me.