Solid volume rotation around y-axis

In summary, the problem involves tracing a pumpkin shape on graph paper and finding its volume when rotated around the y-axis. This can be done by dividing the shape into a semi-circle and a rectangle, and then using the formula v = integral from a to b pi r^2 times thickness (dx or dy) to calculate the volume for each shape. The final answer will be the sum of the volumes of the semi-circle and rectangle. The calculation for the semi-circle involves using the formula of a circle with center h,k: (x-h)^2 + (y-k)^2 = r^2 and then solving for x. The calculation for the rectangle involves simply finding its area and using it as the radius in the volume formula. It
  • #1
ParoXsitiC
58
0

Homework Statement



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:

r5AZB.png


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

Homework Equations



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

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

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.
 
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  • #2
ParoXsitiC said:
We know to be with respect to y there we solve for x:

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

This will give you the total volume



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
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
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, that's 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
Basically what you do is find the volume of the "pumpkin" and subtract the rectange from it.
 
  • #5
The volume of the rectange
 
  • #6
Why would you subtract the rectangle from it? It's apart of the pumpkin.
 
  • #7
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.
 
  • #8
SammyS said:
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.

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.
 

1. What is solid volume rotation around the y-axis?

Solid volume rotation around the y-axis is a mathematical concept that involves rotating a three-dimensional shape around an imaginary line, called the y-axis, to create a new shape with a larger or smaller volume.

2. How is solid volume rotation around the y-axis different from other types of rotation?

Solid volume rotation around the y-axis is different from other types of rotation because it involves rotating a three-dimensional shape as a whole, rather than just rotating a single point or object within the shape. This results in a new shape with a different volume.

3. What is the formula for calculating solid volume rotation around the y-axis?

The formula for calculating solid volume rotation around the y-axis is V = ∫(πr^2)dx, where V is the volume, r is the radius of the shape, and dx is the distance along the y-axis.

4. What are some real-world applications of solid volume rotation around the y-axis?

Solid volume rotation around the y-axis has many real-world applications, such as in manufacturing and construction, where it is used to create and measure the volume of objects. It is also used in engineering and design to create and analyze 3D models.

5. Are there any limitations to solid volume rotation around the y-axis?

Yes, there are limitations to solid volume rotation around the y-axis. It can only be applied to symmetrical shapes, such as cylinders or spheres, and cannot be used on irregular or asymmetrical shapes. Additionally, the formula for calculating volume rotation only works for shapes with a constant radius along the y-axis.

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