Mathematics, 07.07.2020 07:01 DESTROYERMODE2308

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

Answers: 2

Show answers

Answers

In this exercise we have to use the fundamental calculus theorem to calculate the volume of a rotating cylinder, thus we find that:

$V=32\pi$

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 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 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:

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

Integrating we get;

$2\pi [ 16y-4y^2]$

Using the First Fundamental Theorem of Calculus:

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

$2\pi (16) \\V=32\pi$

See more about volume at link

Answer from: bob4964

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: