Find the area bounded by y - axis and x = 4 - y^(2/3)?
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
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 !
Oh, never mind the arithmetic check, we both got the same answer
Thank you, all. :)
To find the area bounded by the y-axis and the curve x = 4 - y^(2/3), we can use integration.
Step 1: Draw the graph of the curve x = 4 - y^(2/3) to visualize the area we are looking for.
Step 2: Determine the range of y-values for which the curve intersects the y-axis. In this case, it is when x = 0, so we have 0 = 4 - y^(2/3). Solving this equation for y, we get y = 4^(3/2) = 8.
Step 3: Set up the integral that represents the area bounded by the curve and the y-axis. We can integrate with respect to y from 0 to 8, since we found that the range of y-values lies between 0 and 8.
The integral would look like this: A = ∫[0,8] (4 - y^(2/3)) dy
Step 4: Evaluate the integral to find the area. Integrate (4 - y^(2/3)) with respect to y, and then substitute the limits of integration.
A = ∫[0,8] (4 - y^(2/3)) dy
Using the power rule for integration, we can integrate y^(2/3) as (3/5) * y^(5/3).
A = [4y - (3/5) * y^(5/3)] evaluated from 0 to 8.
Plugging in the limits of integration:
A = [4(8) - (3/5) * (8)^(5/3)] - [4(0) - (3/5) * (0)^(5/3)]
Simplifying:
A = 32 - (3/5) * 8^(5/3)
Finally, calculate this expression to find the area bounded by the y-axis and the curve x = 4 - y^(2/3).