Evaluating $ \int_0^1 \int_y^1 \sqrt{1+x^2} dx dy $ $$   \int_0^1 \int_y^1 \sqrt{1+x^2} dx dy $$
I've tried switching the order as per fubini theorem to $\int_y^1 \int_0^1 \sqrt{(1+x^2)} dy dx$
and managed to get $\int_y^1 \sqrt{(1+x^2)} dx$ but am stuck there. How do i continue integrating?

Since $y$ does appear in bounds, you can't apply Fubini that way, you must take the domain into account: $\int_0^1 \int_y^1 f(x,y) dx dy = \int_0^1 \int_0^x f(x,y) dy dx$. Then, your $f$ does not depend on $y$, so there is further simplification.
$$\int_0^1 \int_y^1 \sqrt{1+x^2} dx dy = \int_0^1 \int_0^x \sqrt{1+x^2} dy dx$$
$$= \int_0^1 x \sqrt{1+x^2} dx$$
$$= \left[{1 \over 3}(1+x^2)^{3\over 2}\right]_0^1 = \frac{2 \sqrt{2}-1}{3}$$
