Question

how do I set up a double integral for a teardrop shape?

Answer 1

Hmm interesting question! :D

Do you have an equation for the tear drop shape or no?

Answer 2

it gave me x^4 + y^2=x^4

Answer 3

but that doesn't really make sense so i was hoping to find a more general result.. All i have to do is find a double integral formula. thats it

Answer 4

Hmmm if you can write the function in polar, you might be able to set this up without too much trouble.

Our radius would extend from the center of the tear drop out to the surface ( the function ).
\(\Large 0\le r\le f(r,\theta)\)

And our angle theta would extend all the way around.
\(\Large 0\le\theta\le 2\pi\)

\[\Large \int\limits_{\theta=0}^{2\pi}\quad \int\limits_{r=0}^{f(r,\theta)}\quad dr\;d \theta\]Mmmmmm like that maybe? D: hmm