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Over
7 days ago
7

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
Mathematics
1 answer:
zzz [9K]7 days ago
4 0

Answer:

Step-by-step explanation:

If you graph this on a Cartesian coordinate system, it resembles a logarithmic function. It begins at (0, 0) and concludes, based on the boundaries provided, at x = 8. Since we are finding the volume using the shell method, we need to establish x = y equations (which we fortunately have) as well as y intervals. The y-interval is specified in the problem as 0≤y≤2. When we rotate this solid around the line y = 2, since the axis of rotation is above the solid, we need to work backwards to reach there. This is crucial. The formula we apply for shells is

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

where p(y) refers to the distance from the axis of rotation to the solid, and h(y) signifies the height of the rectangle we represent. Notably, since we utilize the shell method, this rectangle is aligned parallel to the axis of rotation, thus, it has a height of 8 units.

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

h(y) represents the rectangle's height, therefore h(y) = 8

and inserting values into the formula gives us:

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

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

Integrating yields

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

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

2\pi(16) which is, finally,

V=32 \pi

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