Volume, cylindrical shell (1 Viewer)

[phyr3]

Curve y = 2 cos 2x in domain 0 (less than or equal to) x (less than or equal to) pi/2. Rotated around x = pi/2.

=X

thanks in adv.

 

[study-freak]

so exactly which area is rotated?

 

[gurmies]

Whole thing, no? Wouldn't they just be two identical areas?

 

[study-freak]

If this is a volumes question as the title suggested, the question must be like "the region bound by the curve y=2cos2x and..... is rotated about the line x=pi/2"

 

[gurmies]

OIC, I haven't done volumes yet, it looked sufficient, but yeah - I see where you're coming from.

 

[study-freak]

yea well, if a line is rotated, it will just be a shell with no volume (It just has a shape as it doesn't even have thickness)
so V=0

 

[phyr3]

ah im sorry , i actually think there is an area. but thats all the question says. bad q i guess.

but i too am confused which area... it's referring to

final answer is: (pi cubed/2) - 2pi

it is also workable via disc method. so it says >.>

 

[study-freak]

I also don't know which area it's referring to just by looking at the answer. I'll have to try different possibilities(there are only two).
I don't have time to do them now though.

 

[Drongoski]

Edit

Note: Integrating by parts, int (2x cos2x) dx = x sin2x + 0.5 cos2x + C

(2 - 2cos2x) comes from (2 - y); 2 itself is the y-intercept of the curve.

 

[phyr3]

tyvm! i got the height wrong =-=

 

[Drongoski]

tyvm!

what's that ?

 

[phyr3]

thank you very much.