## MarcLeclair Group Title Integral question ( finding volume): Find the volume obtained by rotating the following curves bounded by: y=sinx y=0 pi/2≤ x ≤ pi Cylindrical gave me : ∫(pi/2 to pi) 2pi (x) sinxdx but integration by part didn't give me the right answer and the answer key gave me ∫pi(x)^2 dx which I don't understand from where it comes from. one year ago one year ago

1. Outkast3r09 Group Title

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2. Outkast3r09 Group Title

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3. Outkast3r09 Group Title

the depends on how you're rotating it

4. MarcLeclair Group Title

Yeah I rotate it i get an irregularly shape object. Its upon the x axis

5. MarcLeclair Group Title

I have 3 methods for those, disk, cylindrical shell, and washers but I mean I used cylindrical shell in this one.

6. Outkast3r09 Group Title

wait you did it around the x axis?

7. MarcLeclair Group Title

Yeah I thought I mentionned it :/

8. Outkast3r09 Group Title

alright so what did your integral look like before you worked with it

9. Outkast3r09 Group Title

$2\pi\int_{\pi/2}^\pi sin(x)-0dx=2\pi\int_{\pi/2}^\pi sin(x)dx$\]

10. MarcLeclair Group Title

$\int\limits_{\pi/2}^{\pi} 2\pi(x)(sinx) dx = 2\pi \int\limits_{\pi/2}^{\pi} xsin(dx) = -(cosx)(x) - \int\limits_{\pi/2}^{\pi}-cosxdx$

11. MarcLeclair Group Title

I used integration by part with cylindrical shell :/

12. MarcLeclair Group Title

assuming u=x dv= sinx du= dx and v= -cosx

13. agent0smith Group Title

Use the disk method, then you'll have sin^2 x but you can easily integrate that by using the identity for sin^2 x.

14. MarcLeclair Group Title

So can I ask how you differentiate between the cylindrical shell, disk and washer method?

15. agent0smith Group Title

from the disk method, R^2 is sin^2 x, so:$V = \pi \int\limits_{a}^{b} \sin^2 x dx$ and $\sin^2 x = \frac{ 1 - \cos 2x }{2 }$

16. MarcLeclair Group Title

But in this case the cylindrical shell would work.

17. agent0smith Group Title

Use whichever method is easier... disk and washer method are one and the same. The washer method is just a disk with a hole in it, ie the inner radius is zero.

18. MarcLeclair Group Title

should* no?

19. agent0smith Group Title

In this case, cylinder method would involve changing your function into the form y = ... because the cylinders will have to be horizontal to rotate around the x-axis. You're going to have x = arcsin y and so $V = 2 \pi \int\limits_{0}^{1} (y \arcsin y )dy$ ...not worth the hassle when the disk method is ez game.

20. MarcLeclair Group Title

aahhhhh Okay I get it because I have no radius I can use when my function is form of x. Thanks a lot!

21. agent0smith Group Title

|dw:1363166724609:dw| those would be the cylinder cross sections. What i put up there isn't 100% correct for the cylinder method, it should be modified a bit because the cylinders don't touch the y-axis... i think it's more like: $V = 2 \pi \int\limits\limits_{0}^{1} y( \arcsin y - \pi/2)dy$or... something... idk, too much effort to figure it out.