Calculus 3 change of variables

princessp
Messages
4
Reaction score
0

Homework Statement


Use the change of variables to evaluate the integral
(x +y ) sin(x -y )dA, where R is the
region enclosed by y = x, y = x - 2, y = -x and y = -x + 1. (Hint: use u = x + y and
v = x - y

Homework Equations

The Attempt at a Solution


Not sure how to start it
 
Physics news on Phys.org
Draw the region in the xy plane and then draw the region in the uv plane to determine bounds. From there, use the standard change of variables...
 
okay i think that's where i am having trouble. I an not good at drawing the graph
 
princessp said:
okay i think that's where i am having trouble. I an not good at drawing the graph

You have to draw the boundary lines. Don't say that you can not draw the lines y = x, y = x - 2, y = -x, and y = -x + 1 !
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top