[jacks]

Using polar Coordinates , Evaluation of \( \displaystyle \int^{1}_{0}\int^{\sqrt{y}}_{y}\sqrt{x^2+y^2}dxdy \)

[Samir M.]

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.

[delmar]

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

[Samir M.]

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.

[jacks]

Thanks so much friends , Got it.