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

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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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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:
 
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][tex]and[/tex][/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:
 
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
 
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:​
 
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.
 
These are my trials
I have the idea but can't get the right answer
pls reply
THNKX
 

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In your first sheet, you start with [tex]\int A(x)dx[/tex] and then immediately switch to [tex]\int x y[/itex] without a "dx" 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 [tex]y= b\sqrt{1- x^/a^2- z^2/c^2}[/tex] and that [tex]z= c\sqrt{1- x^2/a^2- y^2/b^2[/tex]. The first tells you that when z= 0, [tex]y= b\sqrt{1- x^2/a^2}[/tex] and the second tells you that when y= 0, [tex]z= c\sqrt{1- x^2/a^2}[/tex]. In other words, at each x, the cross section is an ellipse with semi-axes [tex]b\sqrt{1-x^2/a^2}[/tex] and [tex]c\sqrt{1- x^2/a^2}[/tex]. Do you know that the area of an ellipse with semi-axes a and b is [tex]\pi ab[/tex]?[/tex]
 
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: