# Math Help - Area under polar graph

1. ## Area under polar graph

Find the area bounded by:

$r=\frac{1}{\sqrt {1+\theta } }\\$

$0\leqslant \theta \leqslant \pi$

I have so far set up the integral like this:

$\int\nolimits_{0}^{\pi } {d\theta }\int\nolimits_{0}^{\frac{1}{\sqrt {1+\theta } }} {rdr}$

Looks alright?

2. Hello, thepongofping!

Why are you using a double integral?

Find the area bounded by: . $r\:=\:\dfrac{1}{\sqrt {1+\theta}},\;\;0 \le \theta \le\pi$

$\text{Formula: }\;\displaystyle{A \;=\;\tfrac{1}{2}\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta}$

$\displaystyle{\text{We have: }\;A \;=\;\tfrac{1}{2}\!\int^{\pi}_0\! \left(\frac{1}{\sqrt{1+\theta}}\right)^2 d\theta \;=\; \tfrac{1}{2}\!\int^{\pi}_0\frac{d\theta}{1+\theta} }$

. . $A \;=\;\frac{1}{2}\ln|1+\theta|\,\bigg]^{\pi}_0 \;=\;\frac{1}{2}\ln(1+\pi) - \frac{1}{2}\ln(1+0)$

. . $A \;=\;\frac{1}{2}\ln(1+\pi)$