Can you help me , please?

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!

 

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?

 

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.

 

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)?

 

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!

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

 

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

 

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 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}[/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]?

 

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.