I'm trying to solve what seems to be a straightforward double integral but I guess my rewritten integral is wrong. Can this be solved without using a sectant?
My problem: $$ \int_0^{1}\int_{x^2}^x \frac{1}{\sqrt{x^2+y^2}} \; \;dy\;dx $$
Gives:
$$ \int_0^{\pi/4}\int_{0}^1 \frac{1}{r}\cdot r\; \;dr\;d\theta $$
Right? Or did I overlook something?
When I solve that I get $\frac{\pi}{4}$, which is not the correct answer since it should be $\sqrt{2}-1$.
What am I doing wrong?
On
As the comments have pointed out, your domain of integration is incorrect. To solve for the correct one, we have that
$$y = x^2 \implies r\sin\theta = r^2\cos^2\theta$$
which gives the integral
$$I = \int_0^{\frac{\pi}{4}}\int_0^{\tan\theta\sec\theta}1\:dr\:d\theta = \int_0^{\frac{\pi}{4}}\tan\theta\sec\theta\:d\theta = \sec\theta\Bigr|_0^{\frac{\pi}{4}} = \sqrt{2}-1$$
as expected.
This is just an addition to my comment. The required integration domain is the blue region, you calculated the integral on the orange domain (it continues under the blue domain until the $y=x$ curve).