Question

Set up an intergal for the colume of the solid obtained by ratating the region bounded by the parabolas

x=8y-2y^2, x= 4y-y^2

about the x-axis.

Answer 1

I have no idea how to do this :( .

Answer 2

The maximum x-value for each is at y = 2.
The intersections of the two are at (0,0) and (0,4).

Would you like to chop it up into 3 pieces or simply apply Pappus' Theorem?

Answer 3

I can't use Pappus' theorem.

Answer 4

Why not? Specifically proscribed? Not yet presented? Or just not able?

Answer 5

We won't learn it.

Answer 6

@saifoo.khan

Answer 7

First of all, I need to sketch this thing. I can't figure out how the thing looks like.

Answer 8

Think it up yourself and surprise everyone!

3 pieces then.

1) Biggest Piece

x from 4 to 8 and

y from \(2 - \sqrt{\dfrac{8-x}{2}}\) to \(2 + \sqrt{\dfrac{8-x}{2}}\)

2) Top Piece

x from 0 to 4 and

y from \(2 + \sqrt{4-x}\) to \(2 + \sqrt{\dfrac{8-x}{2}}\)

3) Bottom Piece

x from 0 to 4 and

y from \(2 - \sqrt{\dfrac{8-x}{2}}\) to \(2 + \sqrt{4-x}\)

Go!

Answer 9

Why not flip it around to where you are more familiar and solve that problem.

y = 8x-2x^2 and y= 4x-x^2 and wrap it around the y-axis!

Answer 10

But that changes the entire question.

Answer 11

No. The question is to calculate the volume. Unique answers do not care how you find them. It MUST be considered a valid methodology.

Answer 12

Anyway, you can flip it around just to get a look at it, if you can't relate to the present condition.

Answer 13

But if you switch the x and y's that should change the entire shape of the graph.

Answer 14

That's silly. Why would you think that? Absolutely not the case. It is the EXACT SAME problem.

Answer 15

Something like that? REALLY bad sketch though.