Integral Help

[MrCodeDude]

Find the volumes of the solids obtained by rotatiing the region bounded by the curves y = x and y = x^2 about the following lines

a) the x-axis
b) the y-axis
c) y = 2

I really don't get this stuff, so I need help all the way back to square one.

I attempted the first two, got them both wrong, so I figured I'll ask ATOT before I even try the third.

(a) Over the X-Axis (using the disc method)

pi * integral [x^2 - x]^2 from 0 to 1

What's wrong with this setup?

(b)Over the Y-Axis (using the shell method)
2pi * integral [ (x^2 - x)*(x) ] from 0 to 1

What's wrong with this setup?

Thanks a lot in advance. I don't know how I'd be passing half my classes if I didn't have AT 🙂

 

[AFB]

Just fsck it 😀😉

 

[MrCodeDude]

Originally posted by: amdfanboy
Just fsck it 😀😉

Oh how I wish.

 

[MAME]

heh

 

[PowerMacG5]

For (a), the formula is:
pi*int(f(x)^2-g(x)^2)dx from a to b, so for you it's:
pi*int((x^2)^2-x^2)dx from 0 to 1
so it becomes:
pi*int(x^4-x^2)dx from 0 to 1 which I will allow you to solve

and for b, just reverse the x and x^2 so you get:
2*pi*int(x*(x-x^2))dx from 0 to 1.

This is because up until 1, x is greater than x^2.

 

[MrCodeDude]

Originally posted by: Marauder911
For (a), the formula is:
pi*int(f(x)^2-g(x)^2)dx from a to b, so for you it's:
pi*int((x^2)^2-x^2)dx from 0 to 1

Why do you square them individually?

 

[PowerMacG5]

Originally posted by: MrCodeDude

Originally posted by: Marauder911
For (a), the formula is:
pi*int(f(x)^2-g(x)^2)dx from a to b, so for you it's:
pi*int((x^2)^2-x^2)dx from 0 to 1

Why do you square them individually?

Because you aren't talking about disks. You are using washers, and the area of the washer isn't Pi*(f(x)-g(x))^2, it is Pi*(f(x)^2-g(x)^2).

Also look at my edit to the previous post, and you will see your error in your solution for (b).

 

[MrCodeDude]

Originally posted by: Marauder911

Originally posted by: MrCodeDude

Originally posted by: Marauder911
For (a), the formula is:
pi*int(f(x)^2-g(x)^2)dx from a to b, so for you it's:
pi*int((x^2)^2-x^2)dx from 0 to 1

Why do you square them individually?

Because you aren't talking about disks. You are using washers, and the area of the washer isn't Pi*(f(x)-g(x))^2, it is Pi*(f(x)^2-g(x)^2).

Also look at my edit to the previous post, and you will see your error in your solution for (b).

Ahh, thanks 🙂

 

[PowerMacG5]

Originally posted by: MrCodeDude

Originally posted by: Marauder911

Originally posted by: MrCodeDude

Originally posted by: Marauder911
For (a), the formula is:
pi*int(f(x)^2-g(x)^2)dx from a to b, so for you it's:
pi*int((x^2)^2-x^2)dx from 0 to 1

Why do you square them individually?

Because you aren't talking about disks. You are using washers, and the area of the washer isn't Pi*(f(x)-g(x))^2, it is Pi*(f(x)^2-g(x)^2).

Also look at my edit to the previous post, and you will see your error in your solution for (b).

Ahh, thanks 🙂

No problem. AT helps me sometimes when I need integration help, so I can't think of any other way to thank the community other than also helping someone else.

 

[MrCodeDude]

Thanks a lot 🙂

You've been a great help 🙂 :heart:

 

[PowerMacG5]

Originally posted by: MrCodeDude
Thanks a lot 🙂

You've been a great help 🙂 :heart:

No problem. Like I said in my previous post, AT gives me help sometimes with integration, and there is no way better to thank a community then also helping someone else.