Length Q[1]

The length of curve in polar coordinate from $\theta = 0$ to $2\pi$ is given by (see Arc Length of Polar Curves) :

$\begin{aligned} s & = \int_0^{2\pi} \sqrt{r^2 + \left(\frac {dr}{d\theta}\right)^2}\ d\theta & \small \color{#3D99F6} \text{Note that the Cardioid is} \\ & = 2 \int_{-\frac \pi 2}^\frac \pi 2 \sqrt{r^2 + \left(\frac {dr}{d\theta}\right)^2}\ d\theta & \small \color{#3D99F6} \text{symmetrical about the }y\text{-axis} \\ & = 2 \int_{-\frac \pi 2}^\frac \pi 2 \sqrt{(1+\sin \theta)^2 + \cos^2 \theta}\ d\theta \\ & = 2 \int_{-\frac \pi 2}^\frac \pi 2 \sqrt{2+ 2\color{#3D99F6} \sin \theta}\ d\theta & \small \color{#3D99F6} \text{As }\sin x = \cos \left(\frac \pi 2 - x\right) \\ & = 2\sqrt 2 \int_{-\frac \pi 2}^\frac \pi 2 \sqrt{1+ \color{#3D99F6} \cos \left(\frac \pi 2 - \theta\right)}\ d\theta & \small \color{#3D99F6} \text{and }\cos (2x) = 2 \cos^2 x -1 \\ & = 2\sqrt 2 \int_{-\frac \pi 2}^\frac \pi 2 \sqrt{\color{#3D99F6} 2 \cos^2 \left(\frac \pi 4 - \frac \theta 2 \right)}\ d\theta \\ & = 4 \int_{-\frac \pi 2}^\frac \pi 2 \cos \left(\frac \pi 4 - \frac \theta 2 \right)\ d\theta \\ & = - 8 \sin \left(\frac \pi 4 - \frac \theta 2 \right)\bigg|_{-\frac \pi 2}^\frac \pi 2 \\ & = - 8 \left(\sin 0 - \sin \frac \pi 2\right) \\ & = \boxed 8 \end{aligned}$

r is a vector which can be resolved into cartesian components x and y which are dependent on parameter $\theta$ .

$x=r cos(\theta)$ and $y=r cos(\theta)$

$\dfrac{dx}{d\theta}=\dfrac{dr }{d\theta}\times cos(\theta) -r sin\theta$

$\dfrac{dy}{d\theta}=\dfrac{dr }{d\theta}\times sin(\theta) +r cos\theta$

$(\dfrac{dx}{d\theta})^{2}+(\dfrac{dy}{d\theta})^{2} =(\dfrac{dr}{d\theta})^{2} +r^{2}$ [1]

$\sqrt{dx^{2} +dy^{2}}=dL$

$\sqrt{(\dfrac{dx}{d\theta} \times d\theta)^{2} +(\dfrac{dy}{d\theta} \times d\theta)^{2}}=dL$

$\sqrt{(\dfrac{dx}{d\theta})^{2} +(\dfrac{dy}{d\theta})^{2})} \times d\theta=dL$

From[1]

$\sqrt { r^{2} +\dfrac{dr}{d\theta}}\times d\theta=dL$

Taking integral of $dL$ from $\theta =0 to\thinspace 2\pi$ gives L,length of cardioid

$L=\displaystyle \int^{2\pi}_{0}\sqrt { r^{2} +\dfrac{dr}{d\theta}}\times d\theta$

$L=\displaystyle \int^{2\pi}_{0}\sqrt { (1+sin\theta)^{2} +cos^{2}\theta}\times d\theta$

$L=\displaystyle \int^{2\pi}_{0}\sqrt { 2+2sin\theta}\times d\theta =\boxed{8}$