Find the volume of the solid generated by rotating the area bounded by y=x^2 and x=y^2 about the x-axis.
To find the volume of the solid generated by rotating the area bounded by y = x^2 and x = y^2 about the x-axis, we can use the method of cylindrical shells.
First, let's find the intersection points of the curves y = x^2 and x = y^2. Setting the equations equal to each other:
x^2 = y^2
Taking the square root of both sides:
x = ±y
So, the intersection points are (y, y) and (-y, y).
Next, we need to determine the limits of integration. We want to rotate the area bounded by the curves about the x-axis, so we integrate with respect to y from the lower limit (-y) to the upper limit (y).
The height of each cylindrical shell is given by the difference between the two curves: h = y - x^2.
The radius of each cylindrical shell is the distance from the axis of rotation (x-axis) to the curve y = x^2, which is x.
The differential volume of each cylindrical shell is given by dV = 2πrh dy, where r is the radius and h is the height.
Substituting the values for the radius and height, the differential volume becomes dV = 2πxy dy.
To find the total volume, we need to integrate the differential volume over the interval from -y to y:
V = ∫[from -y to y] 2πxy dy
Now, we need to express x in terms of y. From the equations x = ±y, we can choose the positive value x = y.
Substituting x = y into the integral:
V = ∫[from -y to y] 2πy(y) dy
V = 2π ∫[from -y to y] y^2 dy
Evaluating the integral:
V = 2π [y^3/3] (from -y to y)
V = 2π [(y^3/3) - (-y^3/3)]
V = 2π (2y^3/3)
V = (4π/3) y^3
Finally, to find the volume, we need to find the upper limit of integration y.
To determine the intersection points between the curves, we set x = y^2 equal to y = x^2:
y = (y^2)^2
y = y^4
Rearranging the equation:
y^4 - y = 0
Factoring out y:
y(y^3 - 1) = 0
Setting each factor equal to zero:
y = 0 or y^3 - 1 = 0
y = 0 or y = 1
Since we are rotating about the x-axis, the lower limit y = 0.
So, the volume of the solid generated by rotating the area bounded by y = x^2 and x = y^2 about the x-axis is given by:
V = (4π/3) y^3
Setting y = 1 as the upper limit:
V = (4π/3) (1^3)
V = 4π/3
Therefore, the volume of the solid is 4π/3 cubic units.