2 easy double integrals

**coutmath** · May 7th 2012, 12:03 AM

Hello and thanks in advance for your help !

I need those 2 integrals solved until tonight , cause tomorrow morning i'm writing a text and i need them.have tried and failed so far !

View image: MSP7531a185g80d988ih4d000044e7ah94cfgg0fe7
View image: MSP29041a185f2bi3bg12d7000016i6bf8d0i1g1ebi

**DeMath** · May 7th 2012, 02:32 AM

Hint for the second integral: change the order of integration

$\begin{gathered}I= \int\limits_0^1 {\int\limits_0^3 {x\sqrt {{x^2} + {y^2}}\,dydx} } = \int\limits_0^3 {\int\limits_0^1 {x\sqrt {{x^2} + {y^2}}\,dxdy} } \hfill \\[5pt] \left[ \begin{gathered} \sqrt {{x^2} + {y^2}} = t~~ \Rightarrow~~ {x^2} + {y^2} = {t^2}~~ \Rightarrow ~~x\,dx = t\,dt \hfill \\\int {x\sqrt {{x^2} + {y^2}} dx} = \int {{t^2}dt} = \frac{{{t^3}}}{3} + C = \frac{1}{3}{({x^2} + {y^2})^{3/2}} + C \hfill \\ \end{gathered} \right] \hfill \\[5pt] I= \int\limits_0^3 {dy} \left. {\frac{1}{3}{{({x^2} + {y^2})}^{3/2}}} \right|_{x = 0}^{x = 1} = \frac{1}{3}\int\limits_0^3 {\left[ {{{(1 + {y^2})}^{3/2}} - {y^3}} \right]dy=\ldots=\frac{23}{8}\sqrt{10}-\frac{27}{4}-\frac{1}{8}\ln(\sqrt{10}-3)} \hfill \\ \end{gathered}$

**DeMath** · May 7th 2012, 02:56 AM

The first integral

$\begin{aligned}\int\limits_1^3 \int\limits_0^x \frac{2\,dydx}{\sqrt{x^2+y^2}}&= 2\int\limits_1^3 dx\! \left. {\ln\Bigl(y+\sqrt{x^2+y^2}}\Bigr)} \right|_{y = 0}^{y = x} = 2\int\limits_1^3 \Bigl[\ln (x + \sqrt 2 x) - \ln x\Bigr]dx=\\[2pt] &= 2\int\limits_1^3 \Bigl[\ln x + \ln (1 + \sqrt 2 ) - \ln x\Bigr]dx= 4\ln (1 + \sqrt 2 ) \end{aligned}$

**coutmath** · May 7th 2012, 03:09 AM

thank you so much :-))))) ! just solved myself the second one but really thanks for both ! didn't expect such a quick answer !
but concerning first integral my book gives us answer (pi/2)*ln3 ...

**DeMath** · May 7th 2012, 02:08 PM

but concerning first integral my book gives us answer (pi/2)*ln3 ...

I checked in Maple. Check carefully.

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