# Solid of revolution help

• Feb 26th 2011, 09:52 PM
Kuma
Solid of revolution help
Hi.

R is the region enclosed by the x-axis, y-axis and the curve.

y = cos^2(x), x&y>= 0

Find the volume by rotating R about the y-axis.

I'm not sure if I'm doing this right.

So the general formula is:

integral A(x) dx
where A(x) is the area of the strip you're using.

So i graphed the function using the cos^2(x) identity and turning it into 1+cos(2x)/2

anyway after rotating the region it only seemed logical to use horizontal strips since they are circles, the A(x) is given by pi r^2 where r is the curve.

Now my question is, since its a horizontal strip we have to integrate wrt to y. do i need to convert y = 1+cos(2x)/2 in terms of x before i integrate?
• Feb 26th 2011, 10:16 PM
Prove It
First, it helps to know the limits of your integration. In this case, these will be where your graph intersects the axes...

$\displaystyle y$ intercept:

$\displaystyle y = \cos^2{0}$

$\displaystyle = 1$.

So $\displaystyle (0, 1)$ is the $\displaystyle y$ intercept.

$\displaystyle x$ intercept:

$\displaystyle 0 = \cos^2{x}$

$\displaystyle 0 = \cos{x}$

$\displaystyle x = \frac{\pi}{2}$.

So $\displaystyle \left(\frac{\pi}{2}, 0\right)$ is the $\displaystyle x$ intercept.

Since you are rotating around the $\displaystyle y$ axis, it will be a $\displaystyle dy$ integral, so your bounds are $\displaystyle 0 \leq y \leq 1$. Also note that if $\displaystyle y = \cos^2{x}$ then $\displaystyle x = \arccos{\sqrt{y}}$

Now to set up the integral. This solid of revolution can be considered as the area rotated around the $\displaystyle y$ axis. This area needs to be thought of as a series of rectangles. Their length is the same as the value of $\displaystyle x = \arccos{\sqrt{y}}$, and their width is a small change in $\displaystyle y$, which we call $\displaystyle dy$.

When you rotate the rectangles, they form cylinders. The radius of each cylinder is the same as the length of the corresponding rectangle, so the circular cross-sectional area of each cylinder is $\displaystyle \pi \left(\arccos{\sqrt{y}}\right)^2$, and the height of each cylinder is the same as the width of each rectangle.

So the volume of each cylinder is $\displaystyle \pi \left(\arccos{\sqrt{y}}\right)^2\,dy$ and the total volume can be approximated by

$\displaystyle \sum{\pi\left(\arccos{\sqrt{y}}\right)^2\,dy}$.

As you increase the number of cylinders, the sum converges on an integral and the approximation becomes exact. So the volume is

$\displaystyle V = \int_0^1{\pi \left(\arccos{\sqrt{y}}\right)^2\,dy}$.
• Feb 26th 2011, 10:57 PM
Kuma
Hi.
The trouble i mainly had was figuring out how to convert cos^2(x) to make it f(y). Can you please explain how you solve for x with trig functions like that?
• Feb 26th 2011, 10:59 PM
Prove It
$\displaystyle y = \cos^2{x}$

$\displaystyle \sqrt{y} = \cos{x}$

$\displaystyle \arccos{\sqrt{y}} = x$...
• Feb 26th 2011, 11:09 PM
Kuma
Thanks! that makes sense :D