# Find V of a solid made by rotating line on an axis

1. Aug 29, 2011

### TomFoolery

1. The problem statement, all variables and given/known data

If a solid is generated by rotating the line (x=4tan(y*pi/3)) on the y-axis. Find the volume between the area 0≤y≤1.

2. Relevant equations

I know that, when slicing a section (A(x)), I will generate a circle. This gives me two of the dimensions (by using area of a circle formula; A=pi*r2).

r=4tan(y*pi/3)

Therefore, V= $\int$ pi*(4tan(y*pi/3))2
Lower Lim: 0, Upper lim:1

3. The attempt at a solution

This should become:

V = 16*pi $\int$ (tan(y*pi/3)2

Then,
u = tan (y*pi/3)
w = y*pi/3
V = 16*pi $\int$ (u2)(tan(w))(y*pi/3)

then,
V = (16pi) (tan(pi*y/3)3/3) (ln|cos(pi*y/3)|) (pi*y2/6)

My calculator tells me i should end up with 4(3^(1/2) - pi) but I have no idea how. It's just this last step that is killing me.

2. Aug 29, 2011

### lineintegral1

Your integration is not correct. You have the right idea (the radius is the function while the height is dy), but the integral is more elementary than you are making it. Your integral is,
$16\pi\int_{0}^{1}tan^2\frac{y\pi}{3}dy$
Consider a trig identity to make this integral easy to do rather than using a substitution. Let us know if you need anymore help.

3. Aug 29, 2011

### TomFoolery

Got it, it's because Tan2 = sec2(x)-1

this allows you to integrate to get:

(16pi) [(tan(pi*y/3)) (3/pi) -y]

Since the lower limit is 0, the tan part becomes 0 so the answer is F(b) - 0 = F(b)

Thank you for your help, my teacher isn't much assistance so I'm learning how to use the rest of the world as a teacher this semester.

4. Aug 29, 2011

RGV