Find the volume generated by revolving the plane area within
y=2x^2 , y=0 , x=0 , x=5; about x axis
using discs of thickness dx,
v = ∫[0,5] πr^2 dx
where r=y
v = ∫[0,5] π(2x^2)^2 dx
using shells of thickness dy,
v = ∫[0,50] 2πrh dy
where r=y and h = 5-x
v = ∫[0,50] 2πy(5-√(y/2)) dy
To find the volume generated by revolving the plane area within the curves about the x-axis, you can use the method of cylindrical shells.
First, let's find the limits of integration. We want to revolve the region between the curves y = 2x^2 and y = 0 about the x-axis, so we need to find the x-values where these curves intersect.
Setting the two equations equal to each other, we have:
2x^2 = 0
Solving for x, we find x = 0. This is our lower limit of integration.
To find the upper limit of integration, we need to find the x-value where the curve y = 2x^2 intersects with the line x = 5. Plugging x = 5 into the equation, we have:
y = 2(5)^2 = 50
So, the upper limit of integration is x = 5.
Now, let's set up the integral using the method of cylindrical shells. The formula for the volume of a cylindrical shell is:
V = 2π ∫[a, b] x * f(x) * dx
Where a and b are the lower and upper limits of integration, x is the distance from the axis of rotation (in this case, the x-axis), and f(x) is the height of the shell.
In this case, the height of the shell is given by the equation y = 2x^2. Substituting this into the formula, we have:
V = 2π ∫[0, 5] x * (2x^2) * dx
Simplifying the equation, we have:
V = 4π ∫[0, 5] x^3 * dx
Evaluating the integral, we find:
V = 4π * [x^4/4] [0, 5]
V = π * (5^4/4 - 0^4/4)
V = π * (625/4)
Therefore, the volume generated by revolving the plane area within the given curves about the x-axis is π * 625/4 cubic units.