Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^2, x = 4, and y = 1 about the y-axis
39pi/2
225pi/2
735pi/2
None of these
I got 39pi/2
using shells,
v = ∫[1,4] 2πrh dx
where r = x and h = y-1 = x^2-1
v = ∫[1,4] 2π(x)(x^2-1) dx = 225π/2
Hmmm. The area I'm rotating is the shaded triangular region shown at
http://www.wolframalpha.com/input/?i=plot+y%3C%3Dx^2,+16%3E%3Dy%3E%3D1,+0%3C%3Dx%3C%3D4
Using discs,
v = ∫[1,16] π(R^2-r^2) dy
where R=4 and r=√y
v = ∫[1,16] π(16-y) dy = 225π/2
You should have known that 39π/2 is way low, due to the Theorem of Pappus, which states that the volume is the area times the length traveled by the centroid. The area is roughly 15*3/2 = 22, and the centroid would be at about x=3, making a circle of circumference 6π, for a guess of 132π
just did this question and 225 pi/2 is the answer. so what steve said
To find the volume of the solid formed by revolving the region bounded by the given graphs, we can use the method of cylindrical shells.
First, let's find the limits of integration. The region is bounded by y = x^2, x = 4, and y = 1. We need to find the x-values where the curves intersect.
Setting y = x^2 and y = 1 equal to each other, we get:
x^2 = 1
Taking the square root of both sides, we have:
x = ±1
Since we are revolving the region about the y-axis, the limits of integration will be from y = 0 to y = 1.
Now, let's set up the integral to calculate the volume using the cylindrical shell method. The volume of a cylindrical shell is given by the formula V = 2πrhΔy, where r is the radius of the shell, h is the height of the shell, and Δy is the thickness of the shell.
We can consider the x-axis as the radius from the axis of rotation. Therefore, the radius r will be equal to x. The height of the shell h will be equal to the difference between the x-values of the curve y = x^2 and the line x = 4. Thus, h = 4 - x. And the thickness of the shell Δy will be equal to dy since we are integrating with respect to y.
The integral to find the volume is then:
V = ∫[0 to 1] 2π(4 - x)x dy.
Let's evaluate this integral:
V = ∫[0 to 1] 2π(4x - x^2) dy
= 2π ∫[0 to 1] (4x - x^2) dy.
Integrating with respect to y gives:
V = 2π [4xy - (1/3)x^3] evaluated from y = 0 to y = 1.
Now, substitute y = 0 and y = 1 into the integral:
V = 2π [(4x)(1) - (1/3)x^3] evaluated from x = -1 to x = 1.
Evaluating this integral gives:
V = 2π [(4)(1) - (1/3)(1)^3] - 2π [(4)(-1) - (1/3)(-1)^3].
Simplifying:
V = 2π [4 - (1/3)] - 2π [-4 + (1/3)]
= 2π (12/3 - 1/3) - 2π (-12/3 + 1/3)
= 2π (11/3) + 2π (11/3)
= 4π (11/3)
= 44π/3.
Therefore, the correct answer is 44π/3, which is not among the provided options.