Sorry for the badly worded title, my trigonometry knowledge is clearly lacking foundation.
I'm wondering about this because I'm learning how to evaluate double integrals in polar coordinates. How do we know what this integral: $$\int^\sqrt2_0 \int^{4-y^2}_y dxdy $$
would become in polar coordinates?
TLDR: What would the answer be if the upper limit on the original outer integral was $5$ instead of $\sqrt2$?
Detailed thoughts:
It's easy to see that the limits on the inner integral trace out an angle of $\pi/4$, so I have the limits one of the new integrals (the one with respect to $\theta$): the limits are $0$ to $\pi /4$.
But how do I figure out the limits for the radius $r$? In this case I can see that the upper limit on the $y$ integral is $\sqrt2$. If I think of $y$ as being $r$sin$\theta$, then I see that $\sqrt2 / r= $sin$\theta$.
And luckily we know $r=2$ from the upper limit on the $x$ integral, so we see that sin$\theta = \sqrt2/2$ and can conclude that the limits on the outer integral should be $0$ to $1$, giving us our new double integral in polar coordinates:
$$\int_0^1 \int_0^{\pi/4}r \ d\theta \ dr$$
But what if the upper limits on the original outer integral was, for example, $5$ instead of $\sqrt2$?
Your calculations are not accurate.
The two integrals $$\int^\sqrt2_0 \int^{4-y^2}_y dxdy$$ and $$\int_0^1 \int_0^{\pi/4}r \ d\theta \ dr$$
are not the same.
While your $\theta = \pi/4 $ is correct the boundary function $x=4-y^2$ is totally ignored in your calculations.
You need to convert the boundary function into polar form of $r=f(\theta)$ before changing your integralto polar form. coordinvhan