Show that $\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ ?

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Show that $\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ ?

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Answer 1 · 2016-12-19T03:30:03

$\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ QED

Explanation:

We want to evaluate:
$\int \int_{R} \sqrt{x^{2} + y^{2}} d A = \int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x$ $int int _R sqrt(x^2+y^2) dA = int_0^h int_0^x sqrt(x^2+y^2) dy dx$

so here $R$ $R$ is the region:

in the y-direction between $y = 0$ $y=0$ and $y = x$ $y=x$
in the x-direction between $x = 0$ $x=0$ and $x = h$ $x=h$ (a constant)

This represents a right angled triangle in the first quarter.

enter image source here

If we convert to Polar Coordinates then the region $R$ $R$ is:

an angle from $θ = 0$ $theta=0$ to $θ = \frac{π}{4}$ $theta=pi/4$
a ray from $r = 0$ $r=0$ to $r = h sec θ$ $r=hsec theta$ (as $cos θ = \frac{adj}{hyp} = \frac{h}{r}$ $cos theta = "adj"/"hyp"=h/r$ )

And as we convert to Polar coordinates we get:

$x = r cos θ$ $x=rcos theta$
$y = r sin θ$ $y=rsin theta$
$d A = d y d y = r d r d θ$ $dA = dy dy = r dr d theta$

So then the integrand $\sqrt{x^{2} + y^{2}} = \sqrt{r^{2}} = r$ $sqrt(x^2+y^2)=sqrt(r^2)=r$

Hence,
$\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \int_{0}^{\frac{π}{4}} \int_{0}^{h sec θ} r r d r d θ$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = int_0^(pi/4) int_0^(hsec theta) r r dr d theta$
$= \int_{0}^{\frac{π}{4}} \int_{0}^{h sec θ} r^{2} d r d θ$ $" " = int_0^(pi/4) int_0^(hsec theta) r^2 dr d theta$
$= \int_{0}^{\frac{π}{4}} {[\frac{r^{3}}{3}]}_{0}^{h sec θ} d θ$ $" " = int_0^(pi/4) [r^3/3 ]_0^(hsec theta) d theta$
$= \int_{0}^{\frac{π}{4}} (\frac{{(h sec θ)}^{3}}{3} - 0) d θ$ $" " = int_0^(pi/4) ((h sec theta)^3/3 -0) d theta$
$= \frac{h^{3}}{3} \int_{0}^{\frac{π}{4}} ({sec}^{3} θ) d θ$ $" " = h^3/3 int_0^(pi/4) (sec^3 theta) d theta$
$= \frac{h^{3}}{3} {[\frac{1}{2} sec θ tan θ + \frac{1}{2} ln | sec θ + tan θ |]}_{0}^{\frac{π}{4}}$ $" " = h^3/3 [1/2sec theta tan theta + 1/2ln|sec theta + tan theta |]_0^(pi/4)$
$= \frac{h^{3}}{3} {(\frac{1}{2} sec (\frac{π}{4}) tan (\frac{π}{4}) + \frac{1}{2} ln ∣ ∣ sec (\frac{π}{4}) + tan (\frac{π}{4}) ∣ ∣) - (\frac{1}{2} sec 0 tan 0 + \frac{1}{2} ln | sec 0 + tan 0 |)}$ $" " = h^3/3 {(1/2sec (pi/4) tan (pi/4) + 1/2ln|sec (pi/4) + tan (pi/4) |) - (1/2sec 0 tan 0 + 1/2ln|sec 0 + tan 0 |) }$
$= \frac{h^{3}}{3} {(\frac{1}{2} \cdot \sqrt{2} \cdot 1 + \frac{1}{2} ln ∣ ∣ \sqrt{2} + 1 ∣ ∣) - (\frac{1}{2} \cdot 1 \cdot 0 + \frac{1}{2} ln | 1 + 0 |)}$ $" " = h^3/3 {(1/2*sqrt(2) *1 + 1/2ln| sqrt(2) + 1 |) - (1/2*1*0 + 1/2ln|1 + 0 |) }$
$= \frac{h^{3}}{3} (\frac{1}{2} \cdot \sqrt{2} + \frac{1}{2} ln (\sqrt{2} + 1))$ $" " = h^3/3 (1/2*sqrt(2) + 1/2ln( sqrt(2) + 1) )$
$= \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $" " = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ QED

NOTE
As this is an exercise in performing the double integral, I have used (without proof) the results:

$\int {sec}^{3} x d x = \frac{1}{2} sec x tan x + \frac{1}{2} ln | sec x + tan x |$ $int sec^3x dx = 1/2secx tanx + 1/2ln|secx + tanx|$

And for clarity

$sec (\frac{π}{4}) = \sqrt{2}, tan (\frac{π}{4}) = 1, sec 0 = 1, tan 0 = 0$ $sec(pi/4)=sqrt(2), tan(pi/4)=1, sec 0=1, tan 0 =0$

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Show that $\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ ?

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Explanation:

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Show that int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) ) ∫h0∫x0√x2+y2dydx=h36(√2+ln(√2+1)) int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) ) ?

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Show that $\int_{0}^{h} \int_{0}^{x} \sqrt{x^{2} + y^{2}} d y d x = \frac{h^{3}}{6} (\sqrt{2} + ln (\sqrt{2} + 1))$ $int_0^h int_0^x sqrt(x^2+y^2) dy dx = h^3/6 (sqrt(2) + ln( sqrt(2) + 1) )$ ?