
Solid of revolution help
Hi.
The question asks:
R is the region enclosed by the xaxis, yaxis and the curve.
y = cos^2(x), x&y>= 0
Find the volume by rotating R about the yaxis.
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?

First, it helps to know the limits of your integration. In this case, these will be where your graph intersects the axes...
$\displaystyle \displaystyle y$ intercept:
$\displaystyle \displaystyle y = \cos^2{0}$
$\displaystyle \displaystyle = 1$.
So $\displaystyle \displaystyle (0, 1)$ is the $\displaystyle \displaystyle y$ intercept.
$\displaystyle \displaystyle x$ intercept:
$\displaystyle \displaystyle 0 = \cos^2{x}$
$\displaystyle \displaystyle 0 = \cos{x}$
$\displaystyle \displaystyle x = \frac{\pi}{2}$.
So $\displaystyle \displaystyle \left(\frac{\pi}{2}, 0\right)$ is the $\displaystyle \displaystyle x$ intercept.
Since you are rotating around the $\displaystyle \displaystyle y$ axis, it will be a $\displaystyle \displaystyle dy$ integral, so your bounds are $\displaystyle \displaystyle 0 \leq y \leq 1$. Also note that if $\displaystyle \displaystyle y = \cos^2{x}$ then $\displaystyle \displaystyle x = \arccos{\sqrt{y}}$
Now to set up the integral. This solid of revolution can be considered as the area rotated around the $\displaystyle \displaystyle y$ axis. This area needs to be thought of as a series of rectangles. Their length is the same as the value of $\displaystyle \displaystyle x = \arccos{\sqrt{y}}$, and their width is a small change in $\displaystyle \displaystyle y$, which we call $\displaystyle \displaystyle dy$.
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 crosssectional area of each cylinder is $\displaystyle \displaystyle \pi \left(\arccos{\sqrt{y}}\right)^2$, and the height of each cylinder is the same as the width of each rectangle.
So the volume of each cylinder is $\displaystyle \displaystyle \pi \left(\arccos{\sqrt{y}}\right)^2\,dy$ and the total volume can be approximated by
$\displaystyle \displaystyle \sum{\pi\left(\arccos{\sqrt{y}}\right)^2\,dy}$.
As you increase the number of cylinders, the sum converges on an integral and the approximation becomes exact. So the volume is
$\displaystyle \displaystyle V = \int_0^1{\pi \left(\arccos{\sqrt{y}}\right)^2\,dy}$.

Hi.
thanks for the prompt reply!
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?

$\displaystyle \displaystyle y = \cos^2{x}$
$\displaystyle \displaystyle \sqrt{y} = \cos{x}$
$\displaystyle \displaystyle \arccos{\sqrt{y}} = x$...

Thanks! that makes sense :D