how do i use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines
a) y=1/x, y=0, x=1/2, x=2
So I think the integral is 2pi(radius)(height) dx from 0.5 to 2 but I'm not sure what the radius and height would be. the answer is 3pi but i just can't get it no matter how much i try.
the height is the radius.
so y = 1/x ---> r = 1/x
Vol = pi [ integral]r^2 dx from 1/2 to 2
(notice the pi(r^2)dx is your shell or cylinder, with r as the radius, and dx as the width or height of the cylinger)
= pi [integral] 1/x^2 dx from 1/2 to 2
= pi (-1/x) from 1/2 to 2
= pi( -1/2 - (-1/(1/2) )
= pi(-1/2 + 2)
= (p/2)pi
To use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves and lines, follow these steps:
Step 1: Sketch the region
Start by sketching the region bounded by the curves and lines. In this case, the region is bound by the curve y = 1/x, the x-axis (y = 0), the vertical line x = 1/2, and the vertical line x = 2. The region should be in the first quadrant, as y = 1/x is positive in that quadrant.
Step 2: Determine the axis of rotation
The shell method involves rotating the region around an axis to form a solid. In this case, the axis of rotation is the x-axis because the region is bounded vertically by horizontal lines, and we will be integrating with respect to x.
Step 3: Determine the height of the shells
The height of each shell is the difference between the y-values of the two curves that bound the region. In this case, we can see that the upper curve is y = 1/x and the lower curve is y = 0, so the height of each shell is 1/x.
Step 4: Determine the radius of the shells
The radius of each shell is the distance from the axis of rotation (x-axis) to the curve y = 1/x. Since we are integrating with respect to x, the radius is the x-value itself. Therefore, the radius of each shell is x.
Step 5: Set up the integral
To set up the integral for the volume of the solid, use the formula:
Volume = ∫[a, b] 2π(radius)(height) dx,
where [a, b] represents the interval of x-values that bound the region. In this case, the interval is from 1/2 to 2. So the integral becomes:
Volume = ∫[1/2, 2] 2π(x)(1/x) dx.
Step 6: Evaluate the integral
Simplifying the integral, we have:
Volume = ∫[1/2, 2] 2π dx.
Integrating with respect to x, we get:
Volume = 2π[x] from 1/2 to 2,
= 2π(2 - 1/2),
= 2π(3/2),
= 3π.
Thus, the volume of the solid generated by revolving the region bounded by the curves and lines y = 1/x, y = 0, x = 1/2, and x = 2 is 3π.