Integrating a Rhombus: Solving for Area Using the Change of Variables Formula

Bestphysics112
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Homework Statement


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Homework Equations



N/A

The Attempt at a Solution


I solved part a. I got an answer of 140. For part b, however, I am stuck. I came up with a set of points for D in the xy plane [(0,3)(0,6)(4,5)(4,8)] giving me a rhombus. How do i integrate this? I tried to split up the rhombus into two triangles and I still did not get the same answer as a. Where am I going wrong?
 
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Your mapping of ##D## onto ##D^{*}## is a diffeomorphism and hence preserves the boundary and interior. So you need only calculate the action of the diffeomorphism on the boundary of ##D^*##. To calculate the integral just use Fubini's theorem in combination with the change of variables theorem.
 
JonnyG said:
Your mapping of ##D## onto ##D^{*}## is a diffeomorphism and hence preserves the boundary and interior. So you need only calculate the action of the diffeomorphism on the boundary of ##D^*##. To calculate the integral just use Fubini's theorem in combination with the change of variables theorem.
Why would I use the change of variables theorem to compute an integral in the xy plane? Sorry if this sounded too demanding but I've been staring at this problem for quite some time now.
 
Sorry, I didn't see the part that said evaluate directly. But can you show me your work? Make sure you set up your limits of integration carefully.
 
JonnyG said:
Sorry, I didn't see the part that said evaluate directly. But can you show me your work? Make sure you set up your limits of integration carefully.
No problem man! I initially tried setting the points I found as the limits of integration but then realized that I couldn't do that. After I plotted the points I found I saw that they formed a rhombus. So I found an equation for x in terms of y and set it up as follows: ##\intop\nolimits_{3}^{8} \intop\nolimits_{0}^{2y-6} xy dx dy##.

I Evaluated this and I got an answer of 562.5. This is not the same as the answer I obtained using the change of variables formula. What am i doing wrong?
 
Those aren't the correct limits of integration. Look at your parallelogram. x goes from what to what? That should be simple. Now look at the y. It is bounded between two straight lines. What's the equation of those two straight lines?
 
Bestphysics112 said:
No problem man! I initially tried setting the points I found as the limits of integration but then realized that I couldn't do that. After I plotted the points I found I saw that they formed a rhombus. So I found an equation for x in terms of y and set it up as follows: ##\intop\nolimits_{3}^{8} \intop\nolimits_{0}^{2y-6} xy dx dy##
. I Evaluated this and
JonnyG said:
Those aren't the correct limits of integration. Look at your parallelogram. x goes from what to what? That should be simple. Now look at the y. It is bounded between two straight lines. What's the equation of those two straight lines?

x = 2y-6 and x = 2y - 12 ?
 
You might have an easier time if you write the x values varying between 0 and 4, then write the line equations as y = ...
 
JonnyG said:
You might have an easier time if you write the x values varying between 0 and 4, then write the line equations as y = ...
I got the answer! Thank you!.

However, If i wanted to solve it with the integral in dxdy, would i need to use two integrals?
 
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