Question
Find the area bounded by the y-axis and x = 4-y^2/3
Answers
Reiny
I will assume you meant:
x = 4 - y^(2/3)
I assume you made a sketch
let's take horizontal slices ....
the y-intercept is (0,8) so the region is not closed.
I will assume you want the region between the x-axis and and the y-axis
area = ∫4 - y^(2/3) dy from 0 to 8
= 4y - (3/5)y^(5/3) | from 0 to 8
= (32 - (3/5)(8^(5/3)) - 0
= 32 - (3/5)(32)
= 64/5 or 12.8
check my arithmetic
x = 4 - y^(2/3)
I assume you made a sketch
let's take horizontal slices ....
the y-intercept is (0,8) so the region is not closed.
I will assume you want the region between the x-axis and and the y-axis
area = ∫4 - y^(2/3) dy from 0 to 8
= 4y - (3/5)y^(5/3) | from 0 to 8
= (32 - (3/5)(8^(5/3)) - 0
= 32 - (3/5)(32)
= 64/5 or 12.8
check my arithmetic