# Math Help - Parametric Curves and Volume of the area rotated about y-axis

1. ## Parametric Curves and Volume of the area rotated about y-axis

Hi guys! Need help with this question please,

The curve C is defined parametrically by x = (1+t)^(2/3), y = lnt^2, t < than or = to -1
Find the exact volume generated when the area of the region enclosed by C, the lines x = 0, x = 1 and the x-axis is rotated through 2pi radians about the y-axis.

This is my working:
Req. vol. = Vol. of cylinder - Vol. of curve rotated about y axis from y = 0 to y = ln4
(y = ln4 is the intersection b/w C and the line x = 1)
Vol. of cylinder = pi(r^2)(h) = pi(1^2)(ln4) = (pi)(ln4)
For vol. of the curve,
y = ln(t^2) => t = e^(y/2)
Subst. into x = (1+t)^(2/3), x = [1+e^(y/2)]^(2/3)
Vol. of curve = pi * (integrate x^2 dy from y = 0 to y = ln4)
where x^2 = [1+e^(y/2)]^(4/3)

However, the vol. of curve found is much bigger than the vol. of cylinder. What went wrong?

2. ## Re: Parametric Curves and Volume of the area rotated about y-axis

Originally Posted by Blizzardy
Hi guys! Need help with this question please,

The curve C is defined parametrically by x = (1+t)^(2/3), y = lnt^2, t < than or = to -1
Find the exact volume generated when the area of the region enclosed by C, the lines x = 0, x = 1 and the x-axis is rotated through 2pi radians about the y-axis.

This is my working:
Req. vol. = Vol. of cylinder - Vol. of curve rotated about y axis from y = 0 to y = ln4
(y = ln4 is the intersection b/w C and the line x = 1)
Vol. of cylinder = pi(r^2)(h) = pi(1^2)(ln4) = (pi)(ln4)
For vol. of the curve,
y = ln(t^2) => t = e^(y/2)
Subst. into x = (1+t)^(2/3), x = [1+e^(y/2)]^(2/3)
Vol. of curve = pi * (integrate x^2 dy from y = 0 to y = ln4)
where x^2 = [1+e^(y/2)]^(4/3)

However, the vol. of curve found is much bigger than the vol. of cylinder. What went wrong?
you are using cross-sections of disks ... did you look at the graph of the curve? Seems the method of washers is needed to rotate the described region about the y-axis.

also, note that since $t \le -1$ ...

$t^2 = e^y$

$t = -e^{y/2}$

so, $x = (1 - e^{y/2})^{2/3}$

$V = \pi \int_0^{\ln{4}} 1 - (1 - e^{y/2})^{4/3} \, dy$