Question

Exercicios de aprendizagem......

Answer 1

1)
0 < r < 1+2cos(Ф)
2π/3 < Ф <π

$\int\limits^{\pi}_{ \frac{2\pi}{3} } \int\limits_{0}^{ 1+2cos(\theta) }r\;dr\;d\theta\\\\ \frac{1}{2} \int\limits^{\pi}_{ \frac{2\pi}{3} } (1+2cos(\theta))^2 d\theta\\\\ \frac{1}{2} \int\limits^{\pi}_{ \frac{2\pi}{3} } (1+4cos(\theta)+4cos^2(\theta) )d\theta\\\\ \frac{1}{2} \int\limits^{\pi}_{ \frac{2\pi}{3} } (1+4cos(\theta)+4 [\frac{1+cos(2\theta)}{2} ] )d\theta \\\\ \frac{1}{2} \int\limits^{\pi}_{ \frac{2\pi}{3} } (3+4cos(\theta)+2 cos(2\theta) )d\theta$

$\frac{1}{2}*[\left 3\theta+4sen(\theta)+sen(2\theta)\right | ^{\pi}_{\frac{2\pi}{3}}\\\\ \frac{1}{2}*[ (3\pi + 4sen(\pi)+sen(2\pi))-(3*\frac{2\pi}{3}+4sen(\frac{2\pi}{3})+sen(2*\frac{2\pi}{3})) ] \\\\ \frac{1}{2}[3\pi-2\pi -4 \frac{\sqrt{3}}{2}+ \frac{\sqrt{3}}{2} ] = \frac{1}{4}(2\pi-3\sqrt{3} )$

2)
a região será: (area da metade do circulo) -2*(area daquele canto do cardioide)

metade area do circulo = (πr²)/2 = π/2

para o cardioide
0 < r < 1-cos(Ф)
0 < Ф < π/2

$2* \frac{1}{2} \int\limits_{0}^{ \frac{\pi}{2} }(1- cos(\theta))^2 d\theta\\\\ \int\limits_{0}^{ \frac{\pi}{2} }(1-2cos(\theta)+cos^2(\theta))d\theta\\\\ \int\limits_{0}^{ \frac{\pi}{2} }(1-2cos(\theta)+ \frac{1+cos(2\theta)}{2} )d\theta = \left \theta -2sen(\theta)+ \frac{\theta}{2}+ \frac{sen(2\theta)}{4} \right|_{0}^{ \frac{\pi}{2} }\\\\ = ( \frac{\pi}{2}-2+ \frac{\pi}{4})-(0)= \frac{3\pi}{4} -2$

area 
$A= \frac{\pi}{2}- (\frac{3\pi}{4} -2)\\\\A= 2-\frac{\pi}{4}$


3)
$L= 2\int\limits_{0}^{\pi} \sqrt{r^2+( \frac{dr}{d\theta})^2 }\; d\theta\\\\ L= 2\int\limits_{0}^{\pi} \sqrt{(1-cos(\theta))^2+( sen(\theta))^2 }\; d\theta\\\\ L= 2\int\limits_{0}^{\pi} \sqrt{(1-2cos(\theta)+cos^2(\theta)+ sen^2(\theta) }\; d\theta\\\\ L= 2\int\limits_{0}^{\pi} \sqrt{2-2cos(\theta)}\; d\theta\\\\ L=2 \sqrt{2} \int\limits_{0}^{\pi} \sqrt{1-cos(\theta)} \; d\theta$

usando a identidade 
$sen^2(x)=1-cos^2(x)\\\\ \frac{sen(x)}{\sqrt{1+cos(x)}} = \sqrt{1-cos(x)$

$L=2 \sqrt{2} \int\limits_{0}^{\pi} \frac{sen(\theta)}{\sqrt{1+cos(\theta)} } \; d\theta$

U = 1+cos(Ф)
dU = -sen(x) dx

$L=2 \sqrt{2} \int\limits_{0}^{\pi} \frac{-du}{\sqrt{u}} \\\\ L=-2 \sqrt{2} *[2\sqrt{u}] ^{\pi}_0\\\\L=-4\sqrt{2}*[\sqrt{1+cos(\pi)}-\sqrt{1+cos(0)}]\\\\L=-4\sqrt{2}*[0-\sqrt{2}]\\\\L=8$