How do you find the exact length of the polar curve $r = 3 sin (θ)$ $r=3sin(theta)$ on the interval $0 \leq θ \leq \frac{π}{3}$ $0<=theta<=pi/3$ ?

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Answer 1 · 2014-09-21T13:38:54

The arc length is $π$ $pi$ .

Let us look at some details.

$r = 3 sin θ$ $r=3sin theta$

by differentiating with respect to $θ$ $theta$ ,

$\Rightarrow \frac{d r}{d θ} = 3 cos θ$ $Rightarrow {dr}/{d theta}=3cos theta$

So, the arc length L can be found by

$L = \int_{0}^{\frac{π}{3}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $L=int_0^{pi/3}sqrt{r^2+({dr}/{d theta})^2}d theta$

$= \int_{0}^{\frac{π}{3}} \sqrt{3^{2} {sin}^{2} θ + 3^{2} {cos}^{2} θ} d θ$ $=int_0^{pi/3}sqrt{3^2sin^2theta+3^2cos^2theta}d theta$

by pulling $3$ $3$ out of the square-root,

$= 3 \int_{0}^{\frac{π}{3}} \sqrt{{sin}^{2} θ + {cos}^{2} θ} d θ$ $=3int_0^{pi/3}sqrt{sin^2theta+cos^2theta}d theta$

by ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ ,

$= 3 \int_{0}^{\frac{π}{3}} d θ = 3 {[θ]}_{0}^{\frac{π}{3}} = 3 (\frac{π}{3} - 0) = π$ $=3int_0^{pi/3}d theta=3[theta]_0^{pi/3}=3(pi/3-0)=pi$

How do you find the exact length of the polar curve r=3sin(theta)r=3sin(θ)r=3sin(theta) on the interval 0<=theta<=pi/30≤θ≤π30<=theta<=pi/3 ?