Find the area bounded by x=cubed root of y, y=2, y= -1, and y-axis.
To find the area bounded by the given curves, we need to integrate the difference between the upper curve and the lower curve with respect to y.
First, let's find the points of intersection between the curves. We have x = ³√y, y = 2, and y = -1.
Setting x = ³√y equal to y = 2, we can solve for y:
³√y = 2
Cube both sides: y = 8
Setting x = ³√y equal to y = -1, we can solve for y:
³√y = -1
Cube both sides: y = -1
Therefore, the curves intersect at y = 8 and y = -1.
Next, let's determine the upper and lower curves that bound the region. We have x = ³√y and y-axis.
The curve x = ³√y gives the right boundary of the region, while the y-axis gives the left boundary.
To find the area, we will integrate the difference between the upper and lower curves with respect to y.
The integral setup will be as follows:
Area = ∫[from y = -1 to y = 8] [right curve - left curve] dy
The right curve is x = ³√y, and the left curve is x = 0 (y-axis).
Now, let's integrate to find the area:
Area = ∫[y = -1 to y = 8] (³√y - 0) dy
Area = ∫[y = -1 to y = 8] ³√y dy
To solve this integral, we can rewrite it as a power rule with a substitution:
Let u = y^(1/3), then du = (1/3)y^(-2/3) dy.
Substituting in the new variables, the integral becomes:
Area = ∫[u = (-1)^(1/3) to u = 8^(1/3)] u du
Area = ∫[u = -1 to u = 2] u du
Evaluating the integral, we get:
Area = ½[u^2] from -1 to 2
Area = ½[(2)^2 - (-1)^2]
Area = ½(4 - 1)
Area = ½(3)
Area = 1.5
Therefore, the area bounded by the curves x = ³√y, y = 2, y = -1, and the y-axis is 1.5 square units.