Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
x=2y^2
y=1
x=0
about the y-axis
The volume=Pi*Integral(from 0 to 1) (2y^2)^2*dy=Pi(4/5)y^5(0 to 1)=4Pi/5
To find the volume of the solid obtained by rotating the region bounded by the curves, we'll use the method of cylindrical shells.
First, let's graph the region bounded by the curves to get a better understanding of the shape.
The equations x=2y^2, y=1, and x=0 represent a parabolic region that opens towards the right. The y=1 line is a horizontal line at y=1, and x=0 is the y-axis.
Now, let's find the limits of integration. Since we are rotating the region about the y-axis, the limits of integration will be the y-values where the region starts and ends.
From the given equations, we see that y=1 is a horizontal line that bounds the region from below. To find the upper bound, we need to solve the equation x=2y^2 for y. rearranging the equation, we have y^2 = x/2, so y = √(x/2).
Now, let's set up the integral for the volume using the formula for the cylindrical shell:
V = 2π∫(radius)(height)(thickness)dy
The radius of each cylindrical shell is the distance from the y-axis to the curve x=2y^2, which is x=2y^2.
The height of each cylindrical shell is the difference between the upper and lower y-values, which is (1 - √(x/2)).
The thickness of each cylindrical shell is dy.
The integral will be:
V = 2π∫[0 to 1] of [(2y^2)(1 - √(x/2))] dy
Simplifying, we have:
V = 2π∫[0 to 1] of [(2y^2 - 2y√(x/2))] dy
Integrate with respect to y:
V = 2π[(2/3)y^3 - (4/5)(y^2)(√(x/2))] evaluated from 0 to 1
V = 2π[(2/3)(1)^3 - (4/5)(1^2)(√(1/2))] - [(2/3)(0)^3 - (4/5)(0^2)(√(0/2))]
V = 2π[(2/3) - (4/5)(√(1/2))] - 0
V = 2π[(2/3) - (4/5)(√(1/2))]
Now, calculate the final numerical value for the volume using a calculator or approximation.
Therefore, the volume of the solid obtained by rotating the region bounded by the curves x=2y^2, y=1, and x=0 about the y-axis is approximately 1.1804 cubic units.