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
Integrate pi*x^2*dy from y = 0 to y = 1
V = INTEGRAL 4 pi y^4 dy
....(0 to 1)
= (4/5)*pi
To find the volume of the solid obtained by rotating the region bounded by the curves about the y-axis, we can use the method of cylindrical shells.
First, let's graph the region bounded by the curves. The curve x = 2y^2 is a parabola that opens to the right, with its vertex at the origin (0,0). The line y = 1 is a horizontal line passing through the point (0,1). The region bounded by these curves is the region between the parabola and the line, from y = 0 to y = 1.
Now, let's set up the integral to find the volume. The volume of a cylindrical shell with height Δy, radius r, and thickness Δy is given by the formula:
ΔV = 2πrhΔy
In this case, the radius r = x, and the height h = 1 (since the region is bounded by the line y = 1).
To express x in terms of y, we solve the equation x = 2y^2 for x:
x = 2y^2
Now, we can substitute x = 2y^2 and h = 1 into the formula for cylindrical shells:
ΔV = 2π(2y^2)(1)Δy
ΔV = 4πy^2Δy
To find the total volume, we integrate ΔV from y = 0 to y = 1:
V = ∫[0,1] 4πy^2Δy
Evaluating this integral will give us the volume of the solid.