I have a question about this calculus problem. Here is a assignment:
Calculate the area of part of the surface that lays in second quadrant and is bounded by the line $ y = -x$ and by the curves $x^2 + y^2 = 1$ and $x^2 + 3y^2 = 3$.
I know that we have to transform the bounds of the ellipse to the polar coordinates. Problem comes when we try to define radius of the bigger ellipse.
What do you suggest to do?
We have
$$x^2+3y^2=3\iff \frac{x^2}{\left(\sqrt3\right)^2}+\frac{y^2}{1^1}=1$$
and the above is a canonical ellipse with $\;\,x-$ radius $\;\sqrt3\;$ and $\;y\,-$ radius $\;1\;$ . Observe that the intersections are:
$$\begin{cases}x^2+y^2=1\\{}\\ x^2+3y^2=3\end{cases}\implies 2y^2=2\iff y=\pm1\implies x=0$$
so the intersection points of the ellipse and the circle are $\;(0,1),\,\,(0,-1)\;$
And the intersection of the circle and the line:
$$y=-x\implies 1=x^2+y^2=x^2+x^2=2x^2\implies x=\pm\frac1{\sqrt2}=-y$$
and the intersections point in the second quadrant is are $\;\left(-\frac1{\sqrt2}\,,\,\frac1{\sqrt2}\right)\;$ , and the intersection between the line and the ellipse:
$$y=-x\implies 3=x^2+3y^2=x^2+3x^2=4x^2\implies x=\pm\frac{\sqrt3}2=-y$$
and the intersection point in the second quadrant is $\;\left(-\frac{\sqrt3}2\,,\,\frac{\sqrt3}2\right)\;$ , and thus the region is a lunula-like one in the second quadrant, and its area can be calculate by the integrals
$$\int_{-\frac{\sqrt3}2}^{-\frac1{\sqrt2}}(\text{ellipse minus line})dx+\int_{-\frac1{\sqrt2}}^0(\text{ellipse minus circle})dx$$
I leave for you to explicitly write the above and calculate. Take into account that both parts of ellipse and circle we are interested in are in the upper semi-circle (semi-ellipse), so as functions of $\;y\;$ we better choose the poistive square roots...
If you still want the ellipse's parametrization in polar coordinates, try
$$\left(\sqrt3\cos t\,,\,\,\sin t\right)\;,\;\;0\le t\le 2\pi$$
This can work nicely if you want to use double integrals.