Volume of a solid using disks/washers

[toothpaste666]

Homework Statement

Find the volume of the solid generated by rotating the region enclosed by y=\frac{1}{1+x^2} , x=-1,x=1 and y=0 about the line y=2

Homework Equations

pi(outer radius)^2-pi(inner radius)^2

The Attempt at a Solution

Since i am rotating around a horizontal line i figured disks/washers would be a better method than cylindrical shells for this problem. First i sketched the graph. It is sort of an upside down parabola enclosed by the x-axis and x=-1 and x=1. For the outer radius I got 2 and for the inner radius I got 2-\frac{1}{1+x^2}. Since the limits of integration are -1 and 1 and its a parabola I figured I can use symmetry to make it from 0 to 1 and multiply the whole thing by 2.
Step 1:
2\pi \int_0^1(2)^2-(2-\frac{1}{1+x^2})^2\,dx

Step 2:
2\pi \int_0^1(4)-(4-\frac{4}{1+x^2}+\frac{1}{(1+x^2)^2})dx

Step 3:
2\pi \int_0^1\frac{4}{1+x^2}-\frac{1}{(1+x^2)^2}dx

Step 4:
2\pi \int_0^1\frac{4(1+x^2)-1}{(1+x^2)^2}dx

Step 5:
2\pi \int_0^1\frac{4x^2}{(1+x^2)^2}dx

Here I hit a wall because I am not entirely sure how to integrate this. In fact I am not even sure if I am correct up to this point. I would appreciate it greatly if you guys could point me in the right direction.

 

[SammyS]

Homework Statement

Find the volume of the solid generated by rotating the region enclosed by y=\frac{1}{1+x^2} , x=-1,x=1 and y=0 about the line y=2

Homework Equations

pi(outer radius)^2-pi(inner radius)^2

The Attempt at a Solution

Since i am rotating around a horizontal line i figured disks/washers would be a better method than cylindrical shells for this problem. First i sketched the graph. It is sort of an upside down parabola enclosed by the x-axis and x=-1 and x=1. For the outer radius I got 2 and for the inner radius I got 2-\frac{1}{1+x^2}. Since the limits of integration are -1 and 1 and its a parabola I figured I can use symmetry to make it from 0 to 1 and multiply the whole thing by 2.
Step 1:
2\pi \int_0^1(2)^2-(2-\frac{1}{1+x^2})^2\,dx

Step 2:
2\pi \int_0^1(4)-(4-\frac{4}{1+x^2}+\frac{1}{(1+x^2)^2})dx

Step 3:
2\pi \int_0^1\frac{4}{1+x^2}-\frac{1}{(1+x^2)^2}dx

Step 4:
2\pi \int_0^1\frac{4(1+x^2)-1}{(1+x^2)^2}dx

Step 5:
2\pi \int_0^1\frac{4x^2}{(1+x^2)^2}dx

Here I hit a wall because I am not entirely sure how to integrate this. In fact I am not even sure if I am correct up to this point. I would appreciate it greatly if you guys could point me in the right direction.

Hello toothpaste666. Welcome to PF !

It's not a parabola, but it is symmetric w.r.t the y-axis.

4(1 + x²) - 1 = 4x² - 3 ≠ 4x² .

 

[toothpaste666]

Thank you! Wow I can't believe I made that mistake thank you for catching me on that. Wouldnt it be 4X^2+3? The rest of it is right so far though? I was a little unsure about my inner and outer radius and limits of integration. If that is the case I have been having trouble finding a method to integrate that. I played around with a couple of substitutions and partial fractions but I didn't make much progress.

 

[haruspex]

I wouldn't do step 4. After step 3, the left hand integral is well-known. For the other, try the same trig substitution that solves the left hand one.

 

[toothpaste666]

I wouldn't do step 4. After step 3, the left hand integral is well-known. For the other, try the same trig substitution that solves the left hand one.

Oh i see i should have split the integral. I ended up just making it more complicated. So i should let x = tan(theta) on the integral on the right?

 

[haruspex]

Oh i see i should have split the integral. I ended up just making it more complicated. So i should let x = tan(theta) on the integral on the right?

That worked for me.

 

[toothpaste666]

That worked for me.

I got it. Thank you so much!