Question

evaluate the area of an ellipse 4x^2 +y^2 +4 by using double integration

Answer 1

typo?
4x^2 +y^2 =4
perhaps?

Answer 2

i don't understand

Answer 3

4x^2+y^2+4
is not an ellipse
it's not even a graph
can you see that?

Answer 4

it was given in a test
and now i have to do it and submit tmr

Answer 5

they must have meant 4x^2+y^2=4
right? that would be a graph of a ellipse
what you typed is just an expression because it has no equals sign, do you not agree?

Answer 6

well, assuming it is\[4x^2+y^2=4\]I would look at the first quadrant:so it looks like polar coordinates are a nice choice, in which case we have\[D=\left\{ (r,\theta)|0\le r\le1+\cos\theta,0\le\theta\le\frac{\pi}{2} \right\}\]as I said, this is a quarter if the ellipse, so we need to multiply our integral by 4 to get the whole thing\[4\int_{0}^{\frac{\pi}{2}}\int_{0}^{1+\cos\theta}dA=4\int_{0}^{\frac{\pi}{2}}\int_{0}^{1+\cos\theta}rdrd\theta\]to integrate you may need the identity\[\cos^2\theta=\frac{1}{2}(1+\cos(2\theta))\]