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?
 A: 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.
$$\int_0^1 \int_y^1 \sqrt{1+x^2} dx dy = \int_0^1 \int_0^x \sqrt{1+x^2} dy dx$$
$$= \int_0^1 x \sqrt{1+x^2} dx$$
$$= \left[{1 \over 3}(1+x^2)^{3\over 2}\right]_0^1 = \frac{2 \sqrt{2}-1}{3}$$
A: $\displaystyle I=\int \sqrt{1+x^2}dx$
$\displaystyle =\int \sqrt{1+\sinh^2u}\cosh udu$ [Substitute $x=\sinh u, dx=\cosh udu$]
$\displaystyle=\int \cosh^2udu$
$\displaystyle=\frac 1 2 \int (\cosh 2u+1)du$
$\displaystyle=\frac 1 2(\frac 1 2\sinh u+u)+C$
A: Change the order of integration as
$$ \int_0^1 \int_y^1 \sqrt{1+x^2} dx dy= \int_0^1 \int_0^x \sqrt{1+x^2} dy dx.  $$
Now, it is easy to proceed. 
A: Apply the Classical Fubini theorem. I believe the notation below as things become clearer.
$$
\int_{0\leq y\leq 1} \int_{y\leq x\leq 1}\sqrt{1+x^2} \;dx\;dy
=
\iint_{0\leq y\leq x\leq 1} \sqrt{1+x^2} \;d \, A=
\int_{0\leq x\leq 1}\int_{0\leq y\leq x} \sqrt{1+x^2} \;dy\;dx
$$

Then
\begin{align}
\int_{0\leq x\leq 1}\int_{0\leq y\leq x} \sqrt{1+x^2} \;dy\;dx=
&
\int_{0\leq x\leq 1} x \sqrt{1+x^2}\;dx
\\
=
&
\int_{0\leq x\leq 1} (1+x^2)^\prime (1+x^2)^{\frac{1}{2}}\;dx
\\
=
&
\int_{1\leq u\leq 2} u^{\frac{1}{2}}\;du
\end{align}
