Did you make a sketch?
It is easy to solve that they intersect at (4,2) and (4,-2)
We also have a nice symmetry so I will take horizontal slices from y=0 to y = 2 , then double the area
Area = 2∫(y^2 - (2y^2 - 4) ) dy from 0 to 2
= 2∫(4 - y^2 ) dy from 0 to 2
= 2[ 4y - (1/3)y^3] | from 0 to 2
= 2[8 - 8/3 - 0 ]
= 2[ 16/3 ]
= 32/3
check my arithmetic
Find the area of the region bounded by the graph of the equation 2y^2=x+4 and x=y^2
1 answer