# Strange double integral that I only have a vague idea about.

• Jan 19th 2011, 11:31 AM
Undefdisfigure
Strange double integral that I only have a vague idea about.
It's been a while since I've done advanced calculus 1 so I don't remember every technique of integration that I should. I was focusing on linear algebra for a while. I have scanned my hand writing in doing a problem in the book "Multivariable Calculus, James Stewart 6E." I was thinking about integrating by parts but instead tried to use tables and ran into an obstacle (division by 0). I think it's fair to ask how to do this integral as I have made some effort. I have attached my scan to this thread. Note that it is a pdf file and needs to be rotated twice in the view pull down menu because it is upside down. Thanks.
• Jan 19th 2011, 11:41 AM
TheEmptySet
Quote:

Originally Posted by Undefdisfigure
It's been a while since I've done advanced calculus 1 so I don't remember every technique of integration that I should. I was focusing on linear algebra for a while. I have scanned my hand writing in doing a problem in the book "Multivariable Calculus, James Stewart 6E." I was thinking about integrating by parts but instead tried to use tables and ran into an obstacle (division by 0). I think it's fair to ask how to do this integral as I have made some effort. I have attached my scan to this thread. Note that it is a pdf file and needs to be rotated twice in the view pull down menu because it is upside down. Thanks.

So you have the integral

$\displaystyle \int_{0}^{1}\int_{0}^{y}x\sqrt{y^2-x^2}dxdy$

Use the substitution $u=y^2-x^2 \implies du=-2xdx$ and you new limits of integration are $x=0 \implies u=y^2-0^2=y^2$ and
$x=y \implies u=y^2-y^2=0$

Putting this together gives

$\displaystyle \int_{0}^{1} \left( -\frac{1}{2} \int_{y^2}^{0}u^{\frac{1}{2}}du \right)dy$

This should work out nicely from here.
• Jan 19th 2011, 06:36 PM
Undefdisfigure
New limits of integration?
I had trouble integrating when you changed the limits of integration (again I have scanned my work). From what I read in my book you don't have to change the limits in a case like this. Again, it's been a while since I took calculus so there could be some holes in my reasoning, however when I integrated using the same limits of integration I began with I got a simple answer. Check my attached jpg scans. Thanks.
• Jan 19th 2011, 06:59 PM
TheEmptySet
Quote:

Originally Posted by Undefdisfigure
I had trouble integrating when you changed the limits of integration (again I have scanned my work). From what I read in my book you don't have to change the limits in a case like this. Again, it's been a while since I took calculus so there could be some holes in my reasoning, however when I integrated using the same limits of integration I began with I got a simple answer. Check my attached jpg scans. Thanks.

Yes you do not have to change the limits of integration, but it should give the the same answer either way.

If you just look at the inner integral we get

$\displaystyle \int_{0}^{y}x\sqrt{y^2-x^2}dx$

Using the same substitution as above $u=y^2-x^2 \implies du=-2xdx$

$\displaystyle -\frac{1}{2}\int u^{\frac{1}{2}}du=-\frac{1}{3}u^{\frac{3}{2}}=-\frac{1}{3}\left( y^2-x^2\right)^{\frac{3}{2}}\bigg|_{0}^{y}=\frac{1}{3} y^3$

Now if you evaluate the outer integral you get

$\displaystyle \frac{1}{3}\int_{0}^{1}y^3dy=\frac{1}{12}$
• Jan 19th 2011, 07:05 PM
Prove It
Notice that by not changing the limits to values of $\displaystyle u$, you are required to rewrite your answer in terms of $\displaystyle y$ and then substitute the $\displaystyle y$ values. This usually involves more work, since the original substitution was used to make the integrand (and the resulting integral) easier than it originally was...
• Jan 19th 2011, 08:51 PM
Undefdisfigure
Oops and I forgot to put a plus because of the two minuses that together become a plus giving 1/12 as the answer, not -1/12.