Double integral and Jacobian

[viviseraph00]

Homework Statement

The center of a national park is located at (0,0). A special nature preserve is bounded by by straight lines connecting the points A at (3,2), B at (5,1), C at (8,4) and D at (6,5) in a parallelogram. The yearly rainfall at each point is given by RF(x,y)=x^2+xy+y^2 in inches. Transform this into a rectangle in v - r space using the Jacobian theory we studied and determine the following.

1. Find the average rainfall per year for the entire region.

2. Suppose that we desire to constract a weather station at a point in order to report a number on a regular basis that might represent the average rainfall for the entire preserve. Assuming we pick the center of "mass" of the rainfall density function for this point, find the location of the weather station in the (x,y) system

Homework Equations

The Attempt at a Solution

v=x-y
r=x+2y
dv/dx=1
dv/dy=-1
dr/dx=1
dr/dy=2
the jacobian equals 3
x=(1/3)(2v+r)
y=(1/3)(r-v)
thus substituting everything I get...

3*integral [ ((1/3)(2v+r))^2+((1/3)(2v+r))((1/3)(r-v))+((1/3)(r-v))^2]dvdr
with v bounds from 1 to 4 and r bounds from 6 to 17

a (1/9) will factor out to combine with the 3 to give...
(1/3)* integral [(2v+r)^2+(2v+r)(r-v)+(r-v)^2]dvdr

then 7*integral [r^2+rv+v^2]drdv
integrating this double integral and evaluating at the bounds gives...
17521.80

I am guessing that the original dimensions of the preserve parrallelogram was in miles and that the 17521.80 is total inches for the year for the total square mileage of the preserve which is 9 square miles using cos(theta)=(a.b)/[||a||*||b||] and
area = a*b*sin(theta)

really not sure how to procede from here

part 2 the center of the v,r rectangle is (2.5, 11.5)
since
x=(1/3)(2v+r)
y=(1/3)(r-v)
we just plug in the numbers to get
(5.5,3) which is the "center" of the parrallelogram.

I don't why answer is wrong

 

[mighty2000]

Hello,

Thank you for your response.

For part 1, your approach is correct in terms of setting up the Jacobian transformation and finding the average rainfall for the entire region. However, there seems to be a mistake in your integration. The correct integral should be:

(1/9) * integral[(2v+r)^2 + (2v+r)(r-v) + (r-v)^2]dvdr

= (1/9) * integral[4v^2 + 4vr + r^2 + 2v^2 + 2vr + rv - v^2 - vr + r^2]dvdr

= (1/9) * integral[6v^2 + 6r^2 + 6vr]dvdr

= (1/9) * [2v^3 + 2r^2v + 3vr^2] evaluated at v=1 to 4 and r=6 to 17

= (1/9) * [2(4^3 - 1^3) + 2(17^2 - 6^2) + 3(17*17^2 - 6*6^2) - (2(1^3) + 2(6^2) + 3(6*6^2))]

= (1/9) * [1184 + 1521 + 10206 - 114]

= 17521.778 inches per year

For part 2, you are correct in finding the center of the v-r rectangle to be (2.5, 11.5). However, this is not the center of the parallelogram. The center of the parallelogram can be found by taking the average of the x and y coordinates of points A, B, C, and D. This turns out to be (5.5, 3). So, the location of the weather station in the (x,y) system would be (5.5, 3).

I hope this helps. Let me know if you have any further questions.