# Thread: Volume for region rotated around the x-axis

1. ## Volume for region rotated around the x-axis

I just need help setting them up, these are different because they have two y =. What formula would you use? thanks

5) y
= 2x, y = 2, x = 0

6) y
= x2 + 5, y = 2x + 5

2. Originally Posted by Collegeboy110
I just need help setting them up, these are different because they have two y =. What formula would you use? thanks

5) y
= 2x, y = 2, x = 0
The two "y=" is because they are trying to describe to you the region which to rotate.

On a graph, draw the horizontal line at y=2, and shade everything below it.
Next, draw the vertical line at x=0 and shade everything to right (positive).
Next, draw the line y=2x and shade everything above it. This triangular region that results from the overlap of all three shaddings is the region we are rotating.
$\displaystyle \int_{0}^{1} \pi (2^2 - (2x)^2) \, dx$

Hopefully that gets you started. I think if you figure out the correct region you'll probably be able to get going from there. If not, let me know.

3. Can you give me the general formula to use for any problem? and why is 2^2 at the beggining?

question number 6 is ) y = x^2 + 5, y = 2x + 5

would it be the integral of (x^2+5)^2 - (2x+5)^2?

4. Originally Posted by Freemymind
Can you give me the general formula to use for any problem? and why is 2^2 at the beggining?
In general when rotating f(x) about the x-axis we have

$\displaystyle \int_a^b \pi f^2(x) \, dx$

So we are first rotating $\displaystyle f_1(x) = 2$ about the x axis which gives

$\displaystyle \int_a^b \pi 2^2 \, dx$ (*)

We then are subtracting this volume from $\displaystyle f_2(x) = 2x$ rotated about the x-axis
$\displaystyle \int_a^b \pi (2x)^2 \, dx$ (**)

Subtracting (**) from (*) gives the previous result.

5. Originally Posted by Freemymind
question number 6 is ) y = x^2 + 5, y = 2x + 5

would it be the integral of (x^2+5)^2 - (2x+5)^2?
If you graph these functions you will see $\displaystyle y_1=2x+5 > y_2=x^2+5$ on the x-interval $\displaystyle [0,2]$. (set the two y values to be the same and you will find the limits of integration 0 and 2)

So it would be reversed

$\displaystyle \int_{0}^{2} \pi ((y_1)^2 - (y_2)^2) \, dx$

6. Ok so in general when you have two "Y =" functions you will find the greater one then you minus the 2nd one?