- #1
pig
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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?
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?