# Find the volume of the solid.

• December 12th 2010, 07:25 PM
dmbocci
Find the volume of the solid.
I'm stuck on this problem. The answer is probably obvious, but I second guess myself a lot.

Find the volume of the solid below z=sqrt((x-y)(x+4y)), above the parallelogram in the x-y plane with vertices (0,0)(1,1)(5,0)(4,-1).

I have drawn myself a graph to visualize.
Since I havent figured out the way to input stuff on this forum yet, bare with me.

My setup so far is:
triple integral : integral from 0 to 5, integral from -1 to 1, integral from 0 to Z eqn above dzdydx.
The hint given with the problem is change of variables. I've started working it, but since I cant recall what change of variables Im looking to use, my work doesnt seem right so far.
• December 12th 2010, 09:12 PM
Are you familiar with jacobians? You should integrate using a double integral of the form:

$\int_R\int f(x,y)dxdy=\int_S\int f(g(u,v),h(u,v)))\mid\frac{\partial (x,y)}{\partial (u,v)}\mid dudv$

Find the equations of the lines of the parallelogram so that you can find a transformation T from S in (u,v) to R (x,y) such that S is rectangular. Look in your book for some guidance, I'll help you find the region if you get stuck.

The transformation is a function $T(u,v)=(g(u,v),h(u,v))$

EDIT: Sorry, I worded that wrong and fixed it. I orignally mentioned a transformation T from S in (x,y) to R, however S is in the (u,v) plane NOT (x,y)... I hope I didn't cause any confusion...
• December 12th 2010, 09:20 PM
opt
Hi dmbocci,
First, answer the question: what is the domain that you are integrating over?
• December 12th 2010, 10:04 PM
dmbocci
I know of the jacobian, although we didnt spend very much time on it. So basically I'm figuring out this one from scratch. I have an idea though. I have the equations of the lines of the parallelogram. I have y=x and y=(-1/4)x.
So u=x-y and v=-1/4x-y..? Then it gets a bit confusing..
• December 12th 2010, 10:17 PM
Here are the lines:

$y=-\frac{1}{4}x+\frac{5}{4}$
$y=-\frac{1}{4}$
$y=x$
$y=x-5$

Then you can rewrite them:

$4y+x=5$
$4y-x=0$
$y-x=0$
$x-y=5$

Now if you let $u=y-x$ and $v=4y+x$ the lines $u=0$ and $v=5$ are going to be two of the four lines that bound your region in the u-v plane. So you're going to have a simple rectangular region. Solve these equations for x and y to obtain functions $x=g(u,v)$ and $y=h(u,v)$ by solving the u and v equations for x and y, I obtained:

$x=\frac{1}{5}(v-4u)$

$y=\frac{1}{5}(u+v)$

Now your region S should be $-5\leq u\leq 0, 0 \leq v\leq 5$

Does this make sense so far?
• December 12th 2010, 10:22 PM
Quote:

Originally Posted by dmbocci
I know of the jacobian, although we didnt spend very much time on it. So basically I'm figuring out this one from scratch. I have an idea though. I have the equations of the lines of the parallelogram. I have y=x and y=(-1/4)x.
So u=x-y and v=-1/4x-y..? Then it gets a bit confusing..

This would work, there are multiple possibilities. As long as S turns out to be reasonable.
• December 12th 2010, 10:48 PM
The integral I obtained looks like this:

$\int_{-5}^0\int_0^5\sqrt{\frac{1}{5}(-uv)}\mid\frac{\partial (x,y)}{\partial (u,v)}\mid dvdu$

This should work, don't worry about the negative under the square root, this can be taken as $\sqrt{-\frac{1}{5}u}\sqrt{v}$ and the function $\sqrt{-\frac{1}{5}u}$ is defined on $[-5,0]$...

The Jacobian will be a number, I didn't find it, that should be easy enough.

A solution can be found here as well:

Calc Chat Free Solutions

9th edition, 14.8 problem 25...
• December 13th 2010, 07:11 PM
dmbocci
oh ok, I think I see it for the most part. Thank you very much!