Find the volume of the solid obtained by revolving the region

[iRaid]

Homework Statement

Alright well I think I have this right, but my teacher has it done a different way..

Find the volume of the solid obtained by revolving the region bounded by y=x²+1; y=9-x² about y=-1.

Homework Equations

The Attempt at a Solution

So I thought the integral would be:
$$2\pi\int_{0}^{2}(9-x^{2})^{2}-(x^{2}+3)^{2}dx$$
I thought it was x²+3 because you have to rotate it around y=-1 so the inner radius would be 2 added to the 2nd equation.
Oh yeah and I just did the volume from 0 to 2 and multiplied by 2..

Thanks for any help.

 

[Dick]

Homework Statement

Alright well I think I have this right, but my teacher has it done a different way..

Find the volume of the solid obtained by revolving the region bounded by y=x²+1; y=9-x² about y=-1.

Homework Equations

The Attempt at a Solution

So I thought the integral would be:
$$2\pi\int_{0}^{2}(9-x^{2})^{2}-(x^{2}+3)^{2}dx$$
I thought it was x²+3 because you have to rotate it around y=-1 so the inner radius would be 2 added to the 2nd equation.
Oh yeah and I just did the volume from 0 to 2 and multiplied by 2..

Thanks for any help.

That's totally misguided. The inner radius is only 2 at x=0, isn't it? Did you draw a sketch? At an arbitrary value of x the inner radius is (x^2+1)-(-1), isn't it? What's the outer radius?

 

[iRaid]

y=9-x²-(-1) so 10-x² for the outer radius?

But either way his answer is wrong he didn't even take that into account.

 

[Dick]

y=9-x²-(-1) so 10-x² for the outer radius?

But either way his answer is wrong he didn't even take that into account.

If that's the way your teacher did it, then it looks wrong to me. I thought you were showing the way you did it?

 

[Curious3141]

Homework Statement

Alright well I think I have this right, but my teacher has it done a different way..

Find the volume of the solid obtained by revolving the region bounded by y=x²+1; y=9-x² about y=-1.

Homework Equations

The Attempt at a Solution

So I thought the integral would be:
$$2\pi\int_{0}^{2}(9-x^{2})^{2}-(x^{2}+3)^{2}dx$$
I thought it was x²+3 because you have to rotate it around y=-1 so the inner radius would be 2 added to the 2nd equation.
Oh yeah and I just did the volume from 0 to 2 and multiplied by 2..

Thanks for any help.

The steps in solving these problems are very standard:

1) Sketch the curves and highlight the region of interest. Draw in the axis of rotation.

2) Write an expression for the element of volume.

3) Integrate and evaluate the definite integral.

Try going through the steps systematically.

 

[iRaid]

If that's the way your teacher did it, then it looks wrong to me. I thought you were showing the way you did it?

He just subtracted the first equation (squared) from the second equation (squared).
I thought that he did something wrong with it, I don't think he took into account the y=-1 part.

 

[Curious3141]

He just subtracted the first equation (squared) from the second equation (squared).
I thought that he did something wrong with it, I don't think he took into account the y=-1 part.

If he'd added one to each of the y-expressions before squaring then subtracting, then he's taken the axis into account.

 

[Dick]

He just subtracted the first equation (squared) from the second equation (squared).
I thought that he did something wrong with it, I don't think he took into account the y=-1 part.

So what did he show? What you showed is not just subtracting the first equation squared from the second squared. You magically added two to just one of them. Why?

 

[iRaid]

If he'd added one to each of the y-expressions before squaring then subtracting, then he's taken the axis into account.

No he just left them as they are.

 

[Dick]

No he just left them as they are.

Then that would correspond to a rotation around y=0. I don't think that's what you want.

 

[iRaid]

Then that would correspond to a rotation around y=0. I don't think that's what you want.

Ok that's what I thought, thanks lol.