Question

Find the volume of the region R enclosed by the curves y=3sqrt(sin2x) y=0 and the rotated about the x-axis?

Answer 1

is this the curve
$$\Large y = 3 \sqrt{\sin(2x)}$$

Answer 2

yes

Answer 3

that is not enough conditions to bound the region
https://www.desmos.com/calculator/kyfufafhjd

Answer 4

can you take a screenshot of the question

Answer 5

the region is infinite. you can't find the volume of an infinite region

Answer 6

Ok, one minute

Answer 7

yes i see, hmm. but do you see my point about the region,
https://www.desmos.com/calculator/9vnesnm7ar
The region is endless

Answer 8

http://prntscr.com/73gr91

Answer 9

yeah I do understand, I had no problem in using the formula. I had the problem in finding the boundaries

Answer 10

But anyway, thanks for you help!!!!

Answer 11

what does it say after 'and the '

Answer 12

"Find the volume of the region R enclosed by the curves y = 3 sqrt(sin(2x)) y = 0 and
the < missing words> rotated about the x axis ."

Answer 13

that's all it says, theres nothing in the middle

Answer 14

that doesn't make sense grammatically

Answer 15

yeah I know, but that's how my professor wrote it out lol

Answer 16

if you want to salvage the problem, you can assume that the region is bounded by x =0 and x = pi/2. that seems like a good choice.
then do
$$ \Large { \int_{0}^{\pi / 2 } \pi ~ \left(3\sqrt{\sin(2x)} \right)^2~ dx

}

$$

Answer 17

gabriels trumpet suggests that you can find the volume of an infinite region.

Answer 18

\[\int_{1}^{\infty}x^{-1}~dx=ln(\infty)-\ln(1)=\infty \]
\[\int_{1}^{\infty}\pi (x^{-1})^2~dx=-\pi(\infty^{-1}-1^{-1})=\pi \]

Answer 19

as such, if we have a can of paint of volume pi, we can fill the volume of the rotation, but we dont have enough paint to do a cross section of it :)