# Thread: Finding area and volume of graph

1. ## Finding area and volume of graph

Ok...so I'm not sure what to do with this problem at all. If someone could point me in the right direction it would be greatly appreciated.

"Let f and g be the functions given by the equations $f(x) = \sqrt{x})$ and $g(x)=1-\cos(\frac{\pi x}{2})$. Let A be the shaded region in the first quadrant enclosed by the graphs of f and g, and let B be the shaded region in the first quadrant enclosed by the graph of f, the y-axis and the line y = 1.

(a) Find the area of A.
(b) Find the volume of the solid generated when A is rotated about the x-axis.
(c) Find the volume of the solid generated when B is rotated about the y-axis."

So I have a vague idea for (a)...rectangular approximation method? As for (b) and (c)...no idea...

2. Originally Posted by TheMathTham
Ok...so I'm not sure what to do with this problem at all. If someone could point me in the right direction it would be greatly appreciated.

"Let f and g be the functions given by the equations $f(x) = \sqrt{x})$ and $g(x)=1-\cos(\frac{\pi x}{2})$. Let A be the shaded region in the first quadrant enclosed by the graphs of f and g, and let B be the shaded region in the first quadrant enclosed by the graph of f, the y-axis and the line y = 1.

(a) Find the area of A.
(b) Find the volume of the solid generated when A is rotated about the x-axis.
(c) Find the volume of the solid generated when B is rotated about the y-axis."

So I have a vague idea for (a)...rectangular approximation method? As for (b) and (c)...no idea...

I think a lot of your confusion will be aluded by looking at the graphs of these functions.

To find the area of a, you'll need to find, which of the two functions is the top bound, and which is the bottom bound of area A. Then integrate the "top" - "bottom" from their left intersection (x=0) to their right intersection.

To find the volume of a rotated around the x-axis, you will need to integrate with respect to y. The idea is to add up a bunch of discs. The area of a disc is $\pi*r^2$ where $r^2$ is going to be the difference of the squares of your two equations. Integrate this over the length of your region. (You are in essence adding up a bunch of infinitely thin discs).

Same idea for part c, but integrate with respect to x.

3. Well your first step is the area between 2 curves which you can get by taking the equation of the top curve and subtracting from it the bottom curve and then integrating the result from a to b(in your case [0,1] ).
Then you have to pick your favorite method for revolution of solids. Either the disk method or the shell method. Lets do it using the disk method. For your problem you start by squaring the function that makes the top line and then subtracting the square of the function on the bottom. Then put pi on the outside of an integral with that new function we just composed inside and plug and chug.
What we are doing is making an infinite number of nearly flat cylinders and summing up their volumes. Each one of these cylinders has its radius defined by the output of your function. So its the integral of pi r^2 dx where r is the output of your function and dx or dy is the height of your infinitely flat cylinder. So if you are revolving around the y axis you will be doing this with an f(x) = y and if you are revolving around the x axis it should be f(y)=x. In your example it appears that f(x) is on top and g(x) is on the bottom so it would be pi *integral( f(x)^2 - g(x)^2 ) dx around the y axis. do it with f(y) around x axis. I really need to just get faster at using latex.

4. ah ok i think get it now. can you check my answers?
(a).303
(b).858
(c)2.617