Question

The area bounded by the curve y = x2 and the line y = 4 generates various solids of revolution
when rotated as follows.

Answer 1

it as to rotate it about the x and y axis

Answer 2

for the x axis part i integrated from -2 to 2 of x^2

Answer 3

for the y part should i integrate from 0 to 4 of sqrt y

Answer 4

for y part
INT pi * (sqrt y )^2) dy

Answer 5

yes the part is\[V=\pi \int\limits_{0}^{4}(\sqrt{y})^2dy=\pi \int\limits_{0}^{4}ydy=(\pi/2)(16)=8 \pi\]

Answer 6

about the y-axis I mean

Answer 7

yep we agree on that

Answer 8

How do we do the x part though?
\[\pi \int\limits_{-2}^{2}4^2-x^2dx\]right?

Answer 9

yes, that's it\[\pi \int\limits_{-2}^{2}4^2-x^2dx=\pi(16x-x^3/3)\]evaluated from -2 to 2 is\[\pi [(32-8/3)-(-32+8/3)]=\pi (64-16/3)=176 \pi/3\]

Answer 10

wait are you using the washer method for the x part

Answer 11

Yes. The outer radius is y=4 and the inner radius is x^2.

Answer 12

got you

Answer 13

what is it says to rotate about the line y=4