Question

Let f and g be the functions given by f(x) = 2x(1-x) and g(x) = 3(x-1)√x for 0 ≤ x ≤ 1.
Find the volume of the solid generated when the shaded area enclosed by the graphs of f and g is revolved about the horizontal line y = 2.

I used the washer method for this, and input it into my graphing calc and got -6.48..
it's not supposed to be negative so i did something wrong.

Answer 1

The graph of the functions:

Answer 2

sec, won't let me post attached file..

Answer 3

http://i.imgur.com/oRlBJlw.png
this thingy

Answer 4

looks like washer method

Answer 5

yeah it says that he needs to use the washer method

Answer 6

We need your work to find the mistake

Answer 7

nevermind, he said he used the washer LAUGHING OUT LOUD

Answer 8

\[
\pi\int\limits_0^1[g(x)]^2-[f(x)]^2\;dx
\]

Answer 9

well, not quite.

Answer 10

Gotta put in that 2-g / 2-f

Answer 11

Oh, I did that but reversed the order.. i put (2-f(x))^2 - (2-g(x))^2 dx
and what i got is -19.113, forgot to multiply the pi after integrating.

Answer 12

\[
\pi\int\limits_0^1[2-f(x)]^2 - [2-g(x)]^2\;dx
\]

Answer 13

wait what.

Answer 14

basically, it's (outer radius)^2 - (inner radius)^2

Answer 15

I did that already wio, i got the -19.113. is it supposed to be negative then?..

Answer 16

There is no way for the integrand to be negative.

Answer 17

probably you just made some algebraic mistake

Answer 18

https://www.wolframalpha.com/input/?i=pi+integral+%282+-+3%28x-1%29sqrt%28x%29%29%5E2+-+%282+-+2x%281-x%29%29%5E2%2C+x%3D0..1

Answer 19

Oh i was putting this in..
\[{\pi}\int\limits_{0}^{1} (2-(2x(x-1))^2 - (2-(3(x-1)\sqrt{x}))^2 dx\]

Answer 20

use square brackets and {} also so it is less confusing