+322 Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y.
2x+y^2=8 , x=y
got stuck on this one
+129852 2x + y^2 = 8 , x = y
Find the intersection points by subbing the second equation into the first
2x + x^2 = 8 rearrange as
x^2 + 2x - 8 = 0 factor
(x - 2) ( x + 4) = 8 set both factors to 0 and solve for x and we have that x = 2 and x = - 4
Here's the graph : https://www.desmos.com/calculator/jdelptugab
Then the intersection points are ( -4, -4) and (2, 2)
Because of the presence of the y^2 term....this might be easier to integrate in terms of y
Notice that we can write 2x + y^2 = 8 as x = [ 8 - y^2] / 2 = 4 - (1/2)y^2.....and this is the function on the "right" and y = x is the function on the "left"
So we have
2
∫ [ 4 - (1/2)y^2] - y dy =
-4
2
[ 4y - (1/6)y^3 - (1/2)y^2 ] =
-4
[ 4(2) - (1/6)(2^3) - (1/2)2^2 ] - [ 4(-4) - (1/6)(-4)^3 - (1/2)(-4)^2 ] =
[ 8 - 4/3 - 2 ] - [ -16 + 32/3 - 8 ] =
14/3 - [- 40/3] =
54/3 =
18 units^2
