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I'm doing a workout guide about double integrals and I came across an exercise that I could not resolve for a while.

$$\int_0^2\int_1^2 \frac{x}{\sqrt{1+x^2+y^2}} \,\mathrm dx\,\mathrm dy$$

I guess that the easier order of integration is $dxdy$, because if I try to integrate respect to $y$ first, I would have to deal with the integral of a root of $1+x^2+y^2$, while outside of the root there's no $y$.

I tried Integration by substitution (with $u = 1 + x^2 + y^2$ and $du = 2xdx$) to solve the inner integral, but then I felt I couldn't solve the outside integral.

Any hints you can give me?.

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    $\begingroup$ Why don't you try substituting $x=r\cos\theta$ and $y=r\sin\theta$?? $\endgroup$ Mar 28, 2015 at 2:21
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    $\begingroup$ The region is a rectangle, wouldn't the limits be more difficult? $\endgroup$
    – OiciTrap
    Mar 28, 2015 at 2:23
  • $\begingroup$ First do the integration with respect to $x$ as you planned. That is not too hard. The outer integral over $y$ then has terms like $\sqrt{a + y^2}$ for various values of $a$. Look up this integral in your integral table and take it from there. The answer is a bit messy, but there are no essential difficulties. $\endgroup$ Mar 28, 2015 at 2:25

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The antiderivative may be deduced, but it takes a little patience. I prefer to integrate by parts rather than do a trig sub right away.

$$\begin{align}\int dx \sqrt{x^2+a^2} &= x \sqrt{x^2+a^2} - \int dx \frac{x^2}{\sqrt{x^2+a^2}}\\ &= x \sqrt{x^2+a^2} - \int dx \sqrt{x^2+a^2} + a^2 \int \frac{dx}{\sqrt{x^2+a^2}} \end{align}$$

Now a trig sub is less work:

$$\implies 2 \int dx \sqrt{x^2+a^2} = x \sqrt{x^2+a^2} + a^2 \int dt \, \sec{t} $$

or

$$\int dx \sqrt{x^2+a^2} = \frac12 x \sqrt{x^2+a^2} + \frac12 a^2 \log{\left (x+\sqrt{a^2+x^2}\right )} + C$$

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  • $\begingroup$ Thanks, IDK why I didn't think about trigonometric substitution xD. $\endgroup$
    – OiciTrap
    Mar 28, 2015 at 4:29
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$


\begin{align} &\bbox[5px,#ffd]{\int_{0}^{2}\int_{1}^{2} {x \over \root{1 + x^{2} + y^{2}}}\,\dd x\,\dd y} = \int_{0}^{2}\int_{1}^{2} \nabla\times\pars{\root{1 + x^{2} + y^{2}}\,\hat{y}} \cdot\dd\vec{S} \\[5mm] = &\ \oint_{\partial\bracks{\pars{1,2}\times\pars{0,2}}}\,\,\, \root{1 + x^{2} + y^{2}}\,\hat{y}\cdot\dd\vec{r} \\[5mm] = &\ \int_{0}^{2}\root{1 + 2^{2} + y^{2}}\,\dd y + \int_{2}^{0}\root{1 + 1^{2} + y^{2}}\,\dd y \\[5mm] = &\ \underbrace{\int_{0}^{2}\root{5 + y^{2}}\,\dd y} _{\ds{3 + {5 \over 4}\ln\pars{5}}}\ -\ \underbrace{\int_{0}^{2}\root{2 + y^{2}}\,\dd y} _{\ds{\root{6} + \ln\pars{\root{3} + \root{2}}}} \\[5mm] = &\ \bbx{3 - \root{6} + {5 \over 4}\,\ln\pars{5} - \ln\pars{\root{3} + \root{2}}} \approx 1.4161 \\ & \end{align}
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How hard can it be?? I mean once you substitute $x=r\cos\theta$ and $y=r\sin\theta$, all you need to do it find the limits of $r$ and $\theta$.Now $1 \le r\cos\theta \le 2$ and $0 \le r\sin \theta \le 2$. Squaring both sides it is easy to see that $1 \le r \le 2\sqrt{2}$ . Similarly you can find out that $ 0 \le \theta \le tan^{-1}(2)=\theta_{1}$. Then Substituting them in the integral yields $$\int_{0}^{\theta_1} \int_{1}^{2\sqrt{2}}\frac{r\cos\theta}{\sqrt{1+r^2}} rdrd\theta$$

This can be integrated easily.

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  • $\begingroup$ You have turned an integral over a rectangle into an integral over a piece of an annulus. That cannot be right. $\endgroup$ Mar 28, 2015 at 2:51
  • $\begingroup$ Yeah, if we are integrating a rectangle is impossible that $1 \le r \le 2\sqrt{2}$. $\endgroup$
    – OiciTrap
    Mar 28, 2015 at 3:29

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