Find the curve's length over the specified range.&nbsp;r=8+8cos⁡θ&nbsp;on the interval&nbsp;0≤θ≤π

Step 1 The definite integral that represents the arc length given by Arc length =∫αβr2+(drdθ)2dθ =∫0π(8+8cos⁡θ)2+64sin2θdθ =8∫0π1+2cos⁡θ+cos2θ+sin2θdθ =8∫0π2+2cos⁡θdθ =8∫0π4cos2(θ2)dθ =16∫0πcos⁡(θ2)dθ =32sin⁡(θ2)Big∣0π =32

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zeotropojd

Answered question

2021-12-19

Find the curve's length over the specified range. $r = 8 + 8 \cos θ$ on the interval $0 \leq θ \leq π$

Answer & Explanation

Ronnie Schechter

Beginner2021-12-20Added 27 answers

Step 1
The given function is

r = 8 + 8 \cos θ

on the interval

0 \leq θ \leq π

The length of the polar curve is given by

L = \int_{a}^{b} \sqrt{{(r (θ))}^{2} + {(r^{'} (θ))}^{2}} d θ

First, find the derivative:

r^{'} (θ) {(8 \cos (θ) + 8)}^{'} = - 8 \sin (θ)

Step 2
Finally, calculate the integral

L = \int_{0}^{π} \sqrt{{(8 \cos (θ) + 8)}^{2} + {(- 8 \sin (θ))}^{2}} d θ

= \int_{0}^{π} 8 \sqrt{2} \sqrt{\cos (θ) + 1} d θ

= \int_{0}^{π} 8 \sqrt{2} \sqrt{2 \cos^{2} (\frac{θ}{2})} d θ

= \int_{0}^{π} 16 \cos (\frac{θ}{2}) d θ

= 32 (\sin (\frac{θ}{2})) {|_{{(θ = π)}} - 32 (\sin (\frac{θ}{2})) |}_{(θ = 0)}

= 32

Nadine Salcido

Beginner2021-12-21Added 34 answers

Step 1
The definite integral that represents the arc length given by
Arc length

= \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ

= \int_{0}^{π} \sqrt{{(8 + 8 \cos θ)}^{2} + 64 \sin^{2} θ} d θ

= 8 \int_{0}^{π} \sqrt{1 + 2 \cos θ + \cos^{2} θ + \sin^{2} θ} d θ

= 8 \int_{0}^{π} \sqrt{2 + 2 \cos θ} d θ

= 8 \int_{0}^{π} \sqrt{4 \cos^{2} (\frac{θ}{2})} d θ

= 16 \int_{0}^{π} \cos (\frac{θ}{2}) d θ

= 32 \sin (\frac{θ}{2}) B i g ∣_{0}^{π}

= 32

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