consider the area enclosed between the curves f(x)=x^2 and g(x)=4x
what is the volume obtained by revolving the area between these two curves around the line y=20 ?
please help, i don't know what to do
To find the volume obtained by revolving the area between the curves f(x) = x^2 and g(x) = 4x around the line y = 20, you can use the method of cylindrical shells. Here are the steps to find the volume:
1. First, find the points of intersection between the two curves by setting them equal to each other:
x^2 = 4x
Rearrange the equation to get:
x^2 - 4x = 0
Factor out x:
x(x - 4) = 0
So the points of intersection are x = 0 and x = 4.
2. Next, determine the height of each cylindrical shell. The height of a shell is the difference in the y-coordinates of the curves at a particular x-value. In this case, the height of the shell will be (20 - f(x)) - (20 - g(x)) = g(x) - f(x) since y = 20.
Since g(x) = 4x and f(x) = x^2, the height of the shell will be:
h(x) = g(x) - f(x) = 4x - x^2
3. Now, we need to find the radius of each cylindrical shell. The radius is the x-value at a particular point. In this case, the radius is x.
r(x) = x
4. Now, we can use the formula for the volume of a cylindrical shell:
V = ∫(2πrh(x))dx
The integral should be taken from the lower x-value (0) to the upper x-value (4).
V = ∫[0,4] (2πrx)(g(x) - f(x)) dx
Simplify the expression:
V = 2π ∫[0,4] (x)(4x - x^2) dx
5. Integrate the expression with respect to x:
V = 2π ∫[0,4] (4x^2 - x^3) dx
V = 2π [ (4/3)x^3 - (1/4)x^4 ] | [0,4]
6. Evaluate the definite integral:
V = 2π [ (4/3)(4^3) - (1/4)(4^4) ] - 2π [ (4/3)(0^3) - (1/4)(0^4) ]
V = 2π [ (4/3)(64) - (1/4)(256) ] - 0
V = 2π [ (256/3) - (64) ]
V = 2π [ 256/3 - 192/3 ]
V = 2π [ 64/3 ]
V = (128π) / 3
So the volume obtained by revolving the area between the curves f(x) = x^2 and g(x) = 4x around the line y = 20 is (128π) / 3 cubic units.