Volume Formula for Ellipsoid (a,b,c > 0) using Calculus: Slicing and Rotation

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Homework Help Overview

The discussion revolves around deriving a formula for the volume of an ellipsoid defined by the equation (x^2/a^2) + (y^2/b^2) + (z^2/c^2) = 1, where a, b, and c are positive constants. Participants are exploring the application of calculus, specifically the method of volumes by slicing and rotation about an axis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to sketch the ellipsoid and find the equation of the cross-section. There are mentions of calculating the area of the cross-section using the formula A = πab and integrating this area to find the volume. Some participants express confusion over the complexity of the equations encountered during their attempts.

Discussion Status

There is an ongoing exchange of ideas, with some participants providing guidance on how to approach the problem, such as focusing on horizontal cross-sections and ensuring the equations are in standard form. Multiple interpretations of the problem are being explored, and participants are encouraged to share their work for further assistance.

Contextual Notes

Some participants express frustration over the perceived expectation for others to provide complete solutions, while others emphasize the importance of showing work to facilitate help. There are indications of varying levels of understanding and willingness to engage with the problem-solving process.

vip89
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Can you help me , please??

Can you solve this problem by using Calculus l "" Volumes by Slicing and Rotation About an Axis"" ??
Develop a formula for the volume of an ellipsoid of the form

( x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1, a, b, c > 0.

I want the steps , because I know the final answer .
 
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Welcome to PF!

Hi vip89! Welcome to PF! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help you! :smile:
 
I know that I must sketch,find the equation of the cross section , find the area from A=piab and x^2/a^2+y^2/b^2=1
Finally,I must find the definit integral of the area
the final answer will be 4pi/3 abc
 
I don't know how to do this , and how to apply my steps?!
 
Hi vip89! :smile:
vip89 said:
I know that I must sketch,find the equation of the cross section …

ok … we usually take horizontal cross-sections (don't have to) …

in other words, the intersection with a plane z = constant.

So what is the equation (in x and y) if you put z = w (a constant) … ? :smile:

… and then what is the area? :smile:
 
thnkx very much
 
I trid to solve it by this way,but strang things will appear,the equation be more complicated!
 
Show us your work. You can either take a picture, scan it, and upload it, or you can try using LaTeX in between [noparse]and[/noparse] tags.
 
vip89 said:
I trid to solve it by this way,but strang things will appear,the equation be more complicated!

Hi vip89! :smile:

Show us how far you got …

What equation (in x and y) did you get for the horizontal cross-sections when you put z = w (a constant)? :smile:
 
  • #10
vip89 said:
I want the steps , because I know the final answer .
AND WE WANT YOUR WORK!

lol ... funny kid we're not here to do your hw
 
  • #11
Welcome to this thread, rocomath!

oi … rocomath! … that was vip89's very first post (seven days ago)!

… but he's getting the hang of it now! :smile:

If you can't help, just say "welcome!" :smile:
 
  • #12
tiny-tim said:
oi … rocomath! … that was vip89's very first post (seven days ago)!

… but he's getting the hang of it now! :smile:

If you can't help, just say "welcome!" :smile:
sorry :( I've grown less tolerant towards ppl just wanting us to do their hw.
 
  • #13
These are my trials
I have the idea but can't get the right answer
pls reply
THNKX
 

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  • #14
Continue:
 

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  • #15
In your first sheet, you start with \int A(x)dx and then immediately switch to \int x y[/itex] without a &quot;dx&quot; at all. How did that happen? I also cannot see any good reason for writing x as a function of y and z <b>and</b> writing y as a function of x and z.<br /> You are correct that y= b\sqrt{1- x^/a^2- z^2/c^2} and that z= c\sqrt{1- x^2/a^2- y^2/b^2. The first tells you that when z= 0, y= b\sqrt{1- x^2/a^2} and the second tells you that when y= 0, z= c\sqrt{1- x^2/a^2}. In other words, at each x, the cross section is an ellipse with semi-axes b\sqrt{1-x^2/a^2} and c\sqrt{1- x^2/a^2}. Do you know that the area of an ellipse with semi-axes a and b is \pi ab?
 
  • #16
hmm … let's start at the top of your page 3 …
w² = (1 - x²/a² - y²/b²)c²

which I take it is following my suggestion …
tiny-tim said:
So what is the equation (in x and y) if you put z = w (a constant) … ? :smile:

… and then what is the area? :smile:

And then you try to use the formula A = πab for the area of an ellipse.

BUT … that formula only applies if the equation for the ellipse is in the standard form, with nothing but x² and y² on the left and "= 1" on the right.

You must put w² = (1 - x²/a² - y²/b²)c² into that form first.

Then you will get a "new a and b" that are not the same as the original a and b.

ok … rearrange w² = (1 - x²/a² - y²/b²)c² into the standard form … and then find the area. :smile:
 

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