Volume problem: x = y^2, x = y + 6 about the y-axis

[joshswimmer30]

For the problem x=y^2 x=y+6 about the y-axis, I set it up like this:

. . .pi integral (y+6)2-(y^2)^2 dy from -6 to 3

I'm getting a negative answer, which doesnt make sense. Do you know what I did incorrectly?

 

[pka]

The lower limit should be -2 not -6!

 

[joshswimmer30]

why is that??

 

[pka]

The two intersect at (4,-2) & (9,3).
Therefore, integrate from -2 to 3 dy.

 

[joshswimmer30]

why is that??

 

[pka]

Have you drawn the graph?
If not, do so.
Look at the region.

 

[joshswimmer30]

i realize what i did, i dont know why it send that twice--- when i graphed it i just did y= rad x not y=-radx too

 

[galactus]

Here's another way, for fun. More complicated, but it gives you the same thing.

Using shells:

\(\displaystyle \L\\2{\pi}\int_{0}^{9}x(\sqrt{x}-(x-6))dx-2{\pi}\int_{0}^{4}x((6-x)-\sqrt{x})dx\).........[/1]

Check it this way, you'll see it works out the same.

The disk way is much simpler in this case, if you have the wherewithal, it's good to double check another way to see if you get the same thing.

Here's the area your revolving:

If you take this:

and subtract this:

You get the area you're revolving.

That's what [1] does.