Question

Find the area bounded by y - axis and x = 4 - y^(2/3)?

Answer 1

The y-intercepts are at ±8, and using symmetry, you can see that the area is

2∫[0,8] 4-y^(2/3) dy = 128/5

Answer 2

hits y axis at x = 0
y^2/3 = 4
y = 4^3/2
y = +/- 8
vertex at x = 0
so twice integral from y = 0 to y = 8 of
[4 - y^(2/3)] dy

2 [4 y - (3/5) y^5/3] at y = 8 - zero
2 [ 32 - (3/5)(2^3)^(5/3) ]
2 [ 32 - (3/5)(32 ])
64 (2/5)
check my arithmetic !

Answer 3

Oh, never mind the arithmetic check, we both got the same answer

Answer 4

Thank you, all. :)