Can you help me , please?

1. Apr 30, 2008

vip89

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 .

2. Apr 30, 2008

tiny-tim

Welcome to PF!

Hi vip89! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help you!

3. Apr 30, 2008

vip89

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

4. Apr 30, 2008

vip89

I dont know how to do this , and how to apply my steps?!

5. Apr 30, 2008

tiny-tim

Hi vip89!
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) … ?

… and then what is the area?

6. May 2, 2008

vip89

thnkx very much

7. May 6, 2008

vip89

I trid to solve it by this way,but strang things will appear,the equation be more complicated!!

8. May 6, 2008

Tedjn

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.

9. May 7, 2008

tiny-tim

Hi vip89!

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)?

10. May 7, 2008

rocomath

AND WE WANT YOUR WORK!!!

lol ... funny kid we're not here to do your hw

11. May 7, 2008

tiny-tim

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!

If you can't help, just say "welcome!"

12. May 7, 2008

rocomath

sorry :( i've grown less tolerant towards ppl just wanting us to do their hw.

13. May 9, 2008

vip89

These are my trials
I have the idea but cant get the right answer
THNKX

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14. May 9, 2008

vip89

Continue:

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15. May 10, 2008

HallsofIvy

Staff Emeritus
In your first sheet, you start with $$\int A(x)dx$$ and then immediately switch to $$\int x y[/itex] without a "dx" at all. How did that happen? I also cannot see any good reason for writing x as a function of y and z and writing y as a function of x and z. You are correct that [tex]y= b\sqrt{1- x^/a^2- z^2/c^2}$$ 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. May 11, 2008

tiny-tim

hmm … let's start at the top of your page 3 …
which I take it is following my suggestion …
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.