Question

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. x = 8y2, y ≥ 0, x = 8; about y = 2

Answer 1

Answer:

Step-by-step explanation:

If you were to graph this on a coordinate plane, it looks like a log function. It starts at (0, 0) and ends, according to the boundaries given, at x = 8. Because we are to solve for the volume using the shell method, we need x = y equations (which we have, thankfully!) and y intervals. The y-interval is given to us within the problem as 0≤y≤2. If we are to revolve this solid about the line y = 2, the line of rotation is above the solid, so we have to go backwards to get there. That's very important. The formula we need for shells is

$2\pi\int\limits^2_0 {p(y)h(y)} \, dy$

where p(y) is the distance between the axis of rotation and the solid, and h(y) is the height of the representative rectangle. Since we are using the shell method, the representative rectangle is parallel to the axis of rotation, so it is 8 units long (or high).

p(y) is the distance from y = 2 to the solid, so p(y) = 2 - y

h(y) is the length of the representative rectangle, so h(y) = 8

and filling in the formula:

$2\pi\int\limits^2_0 {(2-y)(8)} \, dy$ and simplifying a bit:

$2\pi\int\limits^2_0 {16-8y} \, dy$

Integrating we get

$2\pi[16y-4y^2]$ from 2 to 0. Using the First Fundamental Theorem of Calculus:

$2\pi[(32-16)-(0)]$ simplifies to

$2\pi(16)$ which is, finally,

$V=32 \pi$