Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$ \int_0^1 \int_y^1 \sqrt{1+x^2} dx dy $$

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?

share|improve this question
add comment

4 Answers 4

up vote 1 down vote accepted

Apply the Classical Fubini theorem. I believe the notation below as things become clearer. $$ \int_{0\leq y\leq 1} \int_{y\leq x\leq 1}\sqrt{1+x^2} \;dx\;dy = \iint_{0\leq y\leq x\leq 1} \sqrt{1+x^2} \;d \, A= \int_{0\leq x\leq 1}\int_{0\leq y\leq x} \sqrt{1+x^2} \;dy\;dx $$ Ilustration of double integral

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}

share|improve this answer
    
thanks alot! that diagram really helped me understannd it! –  paradox Mar 20 '13 at 15:50
add comment

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.

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

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

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

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.