Answer to Question #346765 in Calculus for edizon

Question

find the total length of the curve r=4(1-sin theta) form theta = 90 degree to theta= 270 degree

Answer 1

We can find the length of the curve by the next formula:

$l=\int_{ \theta_1}^{ \theta_2} \sqrt{(r(\theta))^2+r '(\theta))^2}d \theta.$

$r(\theta)=4(1-sin\theta),$

$r'(\theta)=-4cos\theta,$

so we have:

$l=\int_{ 90\degree}^{ 270\degree} \sqrt{(4(1-sin\theta))^2+(-4cos\theta)^2}d \theta=$

$=\int_{ 90\degree}^{ 270\degree} \sqrt{16-32sin\theta+16sin^2\theta+16cos^2\theta}d \theta=$

$=\int_{ 90\degree}^{ 270\degree} \sqrt{16-32sin\theta+16}d \theta=$

$=\int_{ 90\degree}^{ 270\degree} \sqrt{32(1-sin\theta)}d \theta=$

$=4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{1-sin\theta}d \theta=$

$=4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{(1-sin\theta) \frac{1+sin\theta}{1+sin\theta}}d \theta=$

$=4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{ \frac{1-sin^2\theta}{1+sin\theta}}d \theta=$

$=4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \sqrt{ \frac{cos^2\theta}{1+sin\theta}}d \theta=$

$=4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \frac{-cos\theta}{\sqrt{{1+sin\theta}}}d \theta=$

$=-4\sqrt{2}\int_{ 90\degree}^{ 270\degree} \frac{cos\theta}{\sqrt{{1+sin\theta}}}d \theta.$

Substitution: $1+sin\theta=u,$ $du=cos\theta d\theta,$

$1+sin(90\degree)=1+1=2,$

$1+sin(270\degree)=1+(-1)=0.$

We have:

$l=-4\sqrt{2}\int_{ 2}^{0} \frac{du}{\sqrt{u}}=4\sqrt{2} \cdot 2\sqrt{u}|_0^2=8\sqrt{2}(\sqrt{2}-0)=16.$

Answer: 16.