# 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?

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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}

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thanks alot! that diagram really helped me understannd it! –  paradox Mar 20 '13 at 15:50

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.

$$\int_0^1 \int_y^1 \sqrt{1+x^2} dx dy = \int_0^1 \int_0^x \sqrt{1+x^2} dy dx$$ $$= \int_0^1 x \sqrt{1+x^2} dx$$ $$= \left[{1 \over 3}(1+x^2)^{3\over 2}\right]_0^1 = \frac{2 \sqrt{2}-1}{3}$$

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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.

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$\displaystyle I=\int \sqrt{1+x^2}dx$

$\displaystyle =\int \sqrt{1+\sinh^2u}\cosh udu$ [Substitute $x=\sinh u, dx=\cosh udu$]

$\displaystyle=\int \cosh^2udu$

$\displaystyle=\frac 1 2 \int (\cosh 2u+1)du$

$\displaystyle=\frac 1 2(\frac 1 2\sinh u+u)+C$

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