1. ## area4

the graph divides the planes into regions .the area of bounded region is

2. ## Re: area4

The graph of the bounded region is rectangle. It has a length of 40, and width of 10.
A = L * W
A = 40 * 10
A = 400.

3. ## Re: area4

how it is a rectangle it can also be a curve

4. ## Re: area4

If you add and subtract $y^2$ you get $x^2+ 2xy+ y^2- y^2+ 40|x|= (x+y)^2- y^2+ |x|= 400$. Let u= x+y and v= y. x= u- y= u- v so the equation becomes $u^2- v^2+ 40|u- v|= 400$.

For x= u- v>0, |x|= |u-v|= u- v so that becomes $u^2+40u- v^2- 40v= 400$. Complete the square by adding 400 and subtracting 400 on the left: $u^2+ 40u+ 400- (v^2+ 40v- 400)= (u+20)^2- (v+20)^2= 400$. That is a hyperbola, in the uv-plane, with center at (-20, -20), vertices (0, -20) and (-40, -20), and vertical asymptotes at 45 degree angles to the axes. That is, one of the asymptotes is the line u= v and, since we require that u- v> 0 here, we have only the branch of the hyperbola on that side.

If u- v< 0, then |u- v|= v- u> 0 and the equation becomes $u^2- 40u- v^2+ 40v= 400$ and, after completing the square, $(u- 20)^2- (v- 20)^2= 400$, a hyperbola with center at (20, 20), vertices (0, 20) and (40, 20), and u= v and u= -v as asymptotes again. Since u< v now, we have the branch of the hyperbola now on the opposite side. Those don't cross: one lies on one side of v= u and the other on the opposite side, so I don't see how they can create a bounded region!

5. ## Re: area4

I plotted $y^2+2xy+40|x|=400$ using mathematica 8. You can also check the graph by clicking here.