# Rotation around y and x axis + shifting the fuction

• Oct 6th 2010, 03:49 PM
Zamzen
Rotation around y and x axis + shifting the fuction
The first problem states to set up a formula for the volume which u get when the area limited by y=X²+1 and y=5 rotates around y=5.

The second problem states to set up a formula for the volume when the area limited by X=2 and y=(8X)^(1/2 rotates around x=2.

My progress on the first part is F(x)=pi*f(x)^2 integrated from -2 to 2 and I realize that I somehow should change the formula of f(x) so it rotates around y=5. And this is where im stuck. I know the answer but I have problems understanding it entirely. I tought that I would just add + 5 in f(x) which is wrong

On the second part I basically arrive at the same problem. I change it to y variable and get the answer F(y)=pi*f(y)^2 integrated from 0 to 4.
. And then I somehow need to change f(y) so it rotates around 2. And I tought that I just add -2 to the formula in y=(8x)^(1/2) which is incorrect.

I appreciate all theoretical help that I can recive. The reason why I have come to my conclusions is because that if y=X² and I want to shift it two steps to the right then I write y=(X-2)² and if I want to shift it 5 steps up I add +5 so I get y=X²+5. But when I tried to apply this to the volume rotation problems I arrive at the wrong answer.

ps. I was trying to add my integrals from a word pad document but it did not work. How do you guys add youre integrals?

• Oct 6th 2010, 04:51 PM
skeeter
Quote:

Originally Posted by Zamzen
The first problem states to set up a formula for the volume which u get when the area limited by y=X²+1 and y=5 rotates around y=5.

method of disks, taking advantage of symmetry ...

$\displaystyle V = 2\pi \int_0^2 [5-(x^2+1)]^2 \, dx$

Quote:

The second problem states to set up a formula for the volume when the area limited by X=2 and y=(8X)^(1/2 rotates around x=2.
method of disks w/r to y ...

$\displaystyle V = \pi \int_0^4 \left(2-\frac{y^2}{8}\right)^2 \, dy$

method of cylindrical shells ...

$\displaystyle V = 2\pi \int_0^2 (2-x) \cdot \sqrt{8x} \, dx$
• Oct 6th 2010, 05:07 PM
Zamzen
Thank you. I can use the cylindrical shell and disc method up until the point where i have to rotate them around a certain point. For example the first one that you solved. When i add the 5 in the function, is it moved upwards with 5 units so it rotates around y = 5?

And on the second problem my textbook says that it should be (y^2)/8 -2. And the cylindrical method you used is very nice.

But i still dont understand the theory behind rotating it around a certain point to 100%. In ex b, u managed to change it so it rotates around x=2 by adding a 2 in the function. And the cylindrical shell method you used with (2-x) suprises me even more. If you can explain the theory how you shift to a point or recomend a website where i can read about it i would be really thankful.
• Oct 6th 2010, 05:24 PM
skeeter
note the red horizontal distance = radius of rotation of the vertical blue strip to form the cylindrical shell ...

what is the distance of that radius in terms of x ?
• Oct 6th 2010, 05:53 PM
Zamzen
r=2-x
x=1
r=1
• Oct 6th 2010, 06:01 PM
skeeter
Quote:

Originally Posted by Zamzen
r=2-x
x=1
r=1

see where the (2-x) came from now?
• Oct 6th 2010, 06:24 PM
Zamzen
yep.
Is it possible to see some problems about rotations around an axis involving around a certain point? i dont have so many in my text book. Or a website that describes it more clearly. Ive been introduced to it recently and i want to understand more about it.
• Oct 6th 2010, 06:39 PM
skeeter
recommend you perform a "google" search ...

calculus volumes of revolution