# Thread: Volume through washer with integration*cough* Calculus 1

1. ## Volume through washer with integration*cough* Calculus 1

Find the volume of the solid obtained by ratating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y = x^4, y = 1; about y = 2

I get that it's a parabola bounded by y = 1... I see that on my graph. And then it's supposed to rotate forward about the x-axis, not sideways right?

But y = 2? So it's 2 unit's shifted upward?

I can't figure this out. Need some

???

2. ## Re: Volume through washer with integration*cough* Calculus 1

No, it's not a parabola. It's a quartic...

3. ## Re: Volume through washer with integration*cough* Calculus 1

The problem says that you're rotating about y=2. So your washer has an inner radius of 1 and an outer radius of 2-x^4. So you integrate from x=-1 to x=1, pi times (outer radius squared - inner radius squared) dx.

- Hollywood

4. ## Re: Volume through washer with integration*cough* Calculus 1

Okay. I was making arithmetic mistakes, woops xD

Thanks

5. ## Re: Volume through washer with integration*cough* Calculus 1

Originally Posted by ILovePizza
Okay. I was making arithmetic mistakes, woops xD

Thanks
That's a very strange response. No one said anything about any arithmetic mistakes.
By the way, since the upper boundary is a horizontal straight line, You can treat it as simplly a disk, from y= 2 to $\displaystyle y= x^4$.

6. ## Re: Volume through washer with integration*cough* Calculus 1

I never showed my work. It's strange you said it was a strange response :|
I had been working on this problem for a long time last night. And was tired by the time I went to sleep. Woke up right now and started working on the problem again. Turns out it was arithmetic mistakes after integrating. I was tired, I couldn't think.

Thanks though