The question is finding the area enclosed by the curves x=$y^2$ and x+2y=8 using both x and y integrals
First I found the limits by letting x=8-2y. This gave the equation $y^2$+2y-8 which gave y=-4 and y=2.
Putting back into equation gives x=4 and x=16
I calculated with y integral being
$$\int_4^2 (8-2y)-y^2 \,$$ ( lower integral is -4, I can't seem to express)
Now I would like to ask how would you calculate using the x integral
I had a think about this and got
but the answers x and y integrals are different so I think theres a mistake somewhere but I don't know what I did wrong.