Solid volume rotation around y-axis

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
which is about 1327.56 pi.

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
which is about 68.3438 pi.

Answer being about 1395.9 pi.

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.

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

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.

Windowmaker

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

Windowmaker

The volume of the rectange

ParoXsitiC

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

SammyS

You're rotating about the y-axis (x=0) not about x = -2.25.


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.

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.