Volume for region rotated around the x-axis

**Collegeboy110** · March 29th 2009, 08:09 PM

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

**n0083** · March 29th 2009, 08:53 PM

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.
$\int_{0}^{1} \pi (2^2 - (2x)^2) \, dx$

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.

**Freemymind** · March 29th 2009, 09:00 PM

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?

**n0083** · March 29th 2009, 09:09 PM

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

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

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

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

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

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

**n0083** · March 29th 2009, 09:18 PM

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 $y_1=2x+5 > y_2=x^2+5$ on the x-interval $[0,2]$ . (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

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

**Freemymind** · March 29th 2009, 11:11 PM

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

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