How do I solve for the area and volume of a cosinusoidal curve?

[pig]

I have the attached graph, a part of a cosinusoidal curve (not sure if this is a correct translation).

I don't have the function, but if I am correct:

y = 3 cos (x*pi/2)

The questions are:

a) Calculate the area between that part of the curve and the x axis.
b) Calculate the volume of the body created by rotating the part of the curve around the y axis.


a) No problem here.

P = 2 * integral from 0 to 1 of 3cos(xpi/2) dx

I got the solution 12/pi.

b) I got this far:

To calculate the volume I need the area of the circles which constitute the body. To get that, I need to express x in terms of y.

y = 3 cos (x*pi/2)
arccos(y/3) = x*pi/2
x = (2/pi)arccos(y/3)

The area is: P = x^2*pi

So the volume is:

V = integral from 0 to 3 of [(2/pi)arccos(y/3)]^2*pi dy

Let I = integral of [(2/pi)arccos(y/3)]^2*pi dy

I = integral of (4/pi)arccos^2(y/3) dy
I = (4/pi) integral of arccos^2(y/3) dy

substitute t = y/3, dt = dy/3, dy=3dt

I = (12/pi) integral of arccos^2(t) dt

I don't know how to integrate this. Anyone able to help?

 

[da_willem]

I believe you're looking for nr. 12 of the following page: http://www.sosmath.com/tables/integral/integ26/integ26.html

 

[pig]

that was fast! thanks a lot :)

by the way, is there a way for me to calculate it myself that i should have thought of?

 

[da_willem]

If you know how to differentiate arccos(x), you could try integration by parts and using some trigonometric identities?!

 

[pig]

The whole problem was that I didn't know how to integrate arccos(x)dx. I didn't immediately think of dividing it into arccos(x)*1dx and using partial integration.. Stupid, I know :)