# Homework Help: Solid volume rotation around y-axis

1. Oct 11, 2011

### ParoXsitiC

1. The problem statement, all variables and given/known data

We have to trace a pumpkin on graph paper and then find it volumne when rotated around the y-axis. Upon doing so we have 2 pieces we can do. One is a semi circle and the other just a rectangle. Refer to this image:

Where we can see the diameter of the circle is 13.5 and the full width of the shape is 9.

2. Relevant equations

(x-h)^2 + (y-k)^2 = r^2

v = integral from a to b pi r^2 times thickness (dx or dy)

3. The attempt at a solution

Know the diameter is 13.5, we know the radium is (13.5 / 2) or 6.75. Using that we can pinpoint the origin of the semi-circle to (2.25,6.75), where 2.25 is the width of the whole shape (9) minus the radius (6.75).

We use the formula of a circle with center h,k:
(x-h)^2 + (y-k)^2 = r^2

Thus:

(x-2.25)^2 + (y-6.75)^2 = (6.75)^2.

We know to be with respect to y there we solve for x:

x = sqrt( (6.75)^2 - (y-6.75)^2) + 2.25

Now we get into rotating around the y-axis. The semi-circle is shifted 2.25 from the y-axis so you add 2.25 to the equation, giving you a radius of sqrt( (6.75)^2 - (y-6.75)^2) + 4.5 thus:

v= pi * integral from 0 to 13.5 (sqrt( (6.75)^2 - (y-6.75)^2) + 4.5)^2 dy

Finally just add to that the retangle rotated around y-axis which is radium 2.25:
v = pi * integral from 0 to 13.5 (2.25)^2 dy

I know this is wrong because if I just do a large rectangle of width 9 and rotate it around I get

v = pi * intefral from 0 to 13.5 (9)^2 dy
or 1093.5 pi. This should be larger that my previous answer, and it's not.

2. Oct 11, 2011

### vrmuth

This will give you the total volume

I don't know why you are doing the above one ,
And you need not find the volume for the rotation of rectangle separately because as per the formula its the volume generated when the AREA UNDER THE CURVE x = sqrt( (6.75)^2 - (y-6.75)^2) + 2.25 from 0 to 13.5 is rotated

3. Oct 11, 2011

### ParoXsitiC

It's my understanding that because the rectangle create a 2.25 by 13.5 gap from the y-axis that when you rotate it around the y-axis it will have a center out hole, thats why you need to add the rectangle to fill that piece.

Also I add 2.25 to the equation because that's the distance from the y-axis.

I believe part of my issue is that I need the semi circle to have a maximum all the way to the right, basically the top of a circle but rotated 90 degrees clockwise. I thought switching x and y around would do that but I am unsure now.

4. Oct 11, 2011

### Windowmaker

Basically what you do is find the volume of the "pumpkin" and subtract the rectange from it.

5. Oct 11, 2011

### Windowmaker

The volume of the rectange

6. Oct 11, 2011

### ParoXsitiC

Why would you subtract the rectangle from it? It's apart of the pumpkin.

7. Oct 11, 2011

### SammyS

Staff Emeritus

Look at the expression, sqrt( (6.75)^2 - (y-6.75)^2) + 2.25 for y = 6.75. you will get 9, which agrees with you diagram. as the radius of the largest horizontal cross-section.

8. Oct 11, 2011

### ParoXsitiC

Ohhh. It makes sense, sorry I am new to the solid volume rotation, but I understand now. You're right.

it would be:

int((sqrt(45.5625-(y-6.75)^2)+2.25)^2, y = 0 .. 13.5) and thus the answer is 800, which makes it fit. Thanks for the explanation.