# Thread: concept behind finding volume between 2 curves

1. ## concept behind finding volume between 2 curves

If there is an area between 2 curves and it's spun around the x-axis I integrate with respect to x, else if it's spun around the y-axis I integrate with respect to y. What screws me up is finding the area because I subtract the wrong function. For example, $\displaystyle y^2=x$ and $\displaystyle 2y=x$. The $\displaystyle y^2$ is on top but the $\displaystyle 2x$ is to the right.
Which rule takes precedence; right-left or top-bottom?

If I get a negative volume, chances are I have the functions switched around, but the converse isn't always true right? For example I could still have the the functions switched around and still get a positive answer?

2. [QUOTE=superdude;340381]If there is an area between 2 curves and it's spun around the x-axis I integrate with respect to x, else if it's spun around the y-axis I integrate with respect to y. What screws me up is finding the area because I subtract the wrong function. For example, $\displaystyle y^2=x$ and $\displaystyle 2y=x$. The $\displaystyle y^2$ is on top but the $\displaystyle 2x$ is to the right.
Which rule takes precedence; right-left or top-bottom?
Are you going from left to right or top to bottom? In other words are you rotating around a vertical line or a horizontal line?

If I get a negative volume, chances are I have the functions switched around, but the converse isn't always true right? For example I could still have the the functions switched around and still get a positive answer?
No. $\displaystyle \int (f(x)- g(x))dx= -\int (g(x)- f(x))dx$. If all you have done is switch f and g, one integral must be the negative of the other. (Unless they are both 0 in which case it doesn't matter!)

3. take the following situations for example:
$\displaystyle f(y)=(y + 4)^2 - 4$ and $\displaystyle g(y)=- y^2 + 6$
is it correct to subtract f(y)-g(y) in accordance to right-left or is it correct to do g(y)-f(y) in accordance of top-bottom?

Am I making myself clear?

another example is where the following functions are rotated around the y-axis:$\displaystyle x=y^2$ and $\displaystyle x=2y$. I keep switching them around doing $\displaystyle \pi\int_0^2 (y^{2})^{2}-(2y)^{2} dy$ which is wrong. However what confuses me is for 0<x<4 $\displaystyle \sqrt{x}>x/2$.

4. Originally Posted by superdude
take the following situations for example:
$\displaystyle f(y)=(y + 4)^2 - 4$ and $\displaystyle g(y)=- y^2 + 6$
is it correct to subtract f(y)-g(y) in accordance to right-left or is it correct to do g(y)-f(y) in accordance of top-bottom?

Am I making myself clear?
Rotating around which axis? Draw a picture! Graph each function and the axis of rotation. Draw a line perpendicular to the axis. Rotated around that axis it becomes a disk. The "outer" and "inner" radii are given by the graphs farther from and closer to the axis of rotation. That determines which is subtracted from which.

another example is where the following functions are rotated around the y-axis:$\displaystyle x=y^2$ and $\displaystyle x=2y$. I keep switching them around doing $\displaystyle \pi\int_0^2 (y^{2})^{2}-(2y)^{2} dy$ which is wrong. However what confuses me is for 0<x<4 $\displaystyle \sqrt{x}>x/2$.
The graphs intersect when $\displaystyle y^2= 2y$. That is, at (0,0) and $\displaystyle (\sqrt{2},2)$.
Since you are rotating around the y-axis, a line perpendicular to the axis (which becomes a disk when rotated) crosses first the the curve $\displaystyle x= y^2$line and then the line x= 2y- that is, the "inner" distance is $\displaystyle y^2$ and the "outer" is 2y. The volume is given by
$\displaystyle \pi \int_{y= 0}^2 (2y)^2- (y^2)^2 dy$

Because you are rotating around the y-axis, the order of y values, $\displaystyle \sqrt{x}$ and x/2 is irrelevant.

5. okay thanks. I thought it was irrelevant which axis the curves are being rotated around but now I see that's not the case. So I've gotta remember to draw a picture and do "outside curve"-"inside curve"

I appreciate the help.
I tried to upload a picture but I don't know how, is there any other way then "insert image"?