### Autor Tema: Double Integration

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02 Febrero, 2022, 11:11 am
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#### jacks

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##### Double Integration
Using polar Coordinates , Evaluation of $\displaystyle \int^{1}_{0}\int^{\sqrt{y}}_{y}\sqrt{x^2+y^2}dxdy$

02 Febrero, 2022, 01:50 pm
Respuesta #1

#### Samir M.

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##### Re: Double Integration
Hi.

The region is limited by $\dfrac{\pi}{4}\le\theta \le \dfrac{\pi}{2}$ and $0\le r\le \dfrac{\cos\theta}{\sin^2\theta}$. The limits for $\theta$ are obvious. For $r$, note that the minimum value is 0 and the maximum value occurs at $y=\sqrt{x} \implies r\sin{\theta} = \sqrt{r\cos{\theta}} \to r = \dfrac{\cos\theta}{\sin^2\theta}$.

Hence, the integral is $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{\cos (\theta)}{\sin ^{2}(\theta)}}r^{2} d r d \theta=\frac{2}{45}(1+\sqrt{2})$.

Regards.
$e^{H_n}=\prod_{k=1}^n e^{1/k}\gt\prod_{k=1}^n\left(1+\frac{1}{k}\right)=n+1 \therefore H_n\gt\log(n+1)$

03 Febrero, 2022, 02:05 am
Respuesta #2

#### delmar

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##### Re: Double Integration
Hello

The region is limited by $0\leq{\theta}\leq{\displaystyle\frac{\pi}{4}}$ and $0\leq{r}\leq{\displaystyle\frac{sen \theta}{cos^2 \theta}}$

Demostrating :

The region R is determinated by $x=y$ and $x=\sqrt[ ]{y}$ then $R=\left\{{(x,y) \ / \ y\leq{x}\leq{\sqrt[ ]{y}}, \ 0\leq{y}\leq{1}}\right\}$ a drawing is convenient

Then $0\leq{\theta}\leq{\displaystyle\frac{\pi}{4}}$ for constant $\theta$,  r varies from zero to point $(x,y) \ / x=\sqrt[ ]{y}\Rightarrow{rcos \theta=\sqrt[ ]{r sen \theta}}\Rightarrow{r=\displaystyle\frac{sen \theta}{cos^2 \theta}}$

Then $\displaystyle\int_{0}^{1}\displaystyle\int_{y}^{\sqrt[ ]{y}}\sqrt[ ]{x^2+y^2} \ dx \ dy=\displaystyle\int_{0}^{\pi/4}\displaystyle\int_{0}^{\displaystyle\frac{sen \theta}{cos^2 \theta}}r^2 \ dr \ d \theta$

Regards

03 Febrero, 2022, 04:42 am
Respuesta #3

#### Samir M.

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##### Re: Double Integration
Hello

The region is limited by $0\leq{\theta}\leq{\displaystyle\frac{\pi}{4}}$ and $0\leq{r}\leq{\displaystyle\frac{sen \theta}{cos^2 \theta}}$

Demostrating :

The region R is determinated by $x=y$ and $x=\sqrt[ ]{y}$ then $R=\left\{{(x,y) \ / \ y\leq{x}\leq{\sqrt[ ]{y}}, \ 0\leq{y}\leq{1}}\right\}$ a drawing is convenient

Then $0\leq{\theta}\leq{\displaystyle\frac{\pi}{4}}$ for constant $\theta$,  r varies from zero to point $(x,y) \ / x=\sqrt[ ]{y}\Rightarrow{rcos \theta=\sqrt[ ]{r sen \theta}}\Rightarrow{r=\displaystyle\frac{sen \theta}{cos^2 \theta}}$

Then $\displaystyle\int_{0}^{1}\displaystyle\int_{y}^{\sqrt[ ]{y}}\sqrt[ ]{x^2+y^2} \ dx \ dy=\displaystyle\int_{0}^{\pi/4}\displaystyle\int_{0}^{\displaystyle\frac{sen \theta}{cos^2 \theta}}r^2 \ dr \ d \theta$

That's assuming the op is integrating w.r.t to $x$ first and then wrt $y$. I've assumed the opposite. Anyway, both regions are congruent and hence the integral value is the same.

Regards.
$e^{H_n}=\prod_{k=1}^n e^{1/k}\gt\prod_{k=1}^n\left(1+\frac{1}{k}\right)=n+1 \therefore H_n\gt\log(n+1)$

03 Febrero, 2022, 03:25 pm
Respuesta #4

#### jacks

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##### Re: Double Integration
Thanks so much friends , Got it.