How do you find the area of the region bounded by the polar curves $r = \sqrt{3} cos (θ)$ $r=sqrt(3)cos(theta)$ and $r = sin (θ)$ $r=sin(theta)$ ?

Question

How do you find the area of the region bounded by the polar curves $r = \sqrt{3} cos (θ)$ $r=sqrt(3)cos(theta)$ and $r = sin (θ)$ $r=sin(theta)$ ?

1 Answer

Impact of this question

20127 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-25T22:49:41

The area of the enclosed region is $\frac{5}{24} π - \frac{\sqrt{3}}{4}$ $5/24 pi-sqrt{3}/4$ .

It looks like this:

enter image source here

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

$= \int_{0}^{\frac{π}{3}} {[\frac{r^{2}}{2}]}_{0}^{sin θ} d θ + \int_{\frac{π}{3}}^{\frac{π}{2}} {[\frac{r^{2}}{2}]}_{0}^{\sqrt{3} cos θ} d θ$ $=int_0^{pi/3} [r^2/2]_0^{sin theta} d theta + int_{pi/3}^{pi/2}[r^2/2]_0^{sqrt{3}cos theta} d theta$

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

$= \frac{1}{4} \int_{0}^{\frac{π}{3}} (1 - sin 2 θ) d θ + \frac{3}{4} \int_{\frac{π}{3}}^{\frac{π}{2}} (1 + cos 2 θ) d θ$ $=1/4int_0^{pi/3} (1-sin2 theta) d theta + 3/4int_{pi/3}^{pi/2}(1+cos2 theta) d theta$

$= \frac{1}{4} {[θ - \frac{sin 2 θ}{2}]}_{0}^{\frac{π}{3}} + \frac{3}{4} {[θ + \frac{sin 2 θ}{2}]}_{\frac{π}{3}}^{\frac{π}{2}}$ $=1/4[theta-{sin2theta}/2]_0^{pi/3} +3/4[theta+{sin2theta}/2]_{pi/3}^{pi/2}$

$= \frac{1}{4} (\frac{π}{3} - \frac{\sqrt{3}}{4}) + \frac{3}{4} (\frac{π}{6} - \frac{\sqrt{3}}{4})$ $=1/4(pi/3-sqrt{3}/4)+3/4(pi/6-sqrt{3}/4)$

$= \frac{5}{24} π - \frac{\sqrt{3}}{4}$ $=5/24 pi-sqrt{3}/4$

Science

Math

Humanities

... and beyond

How do you find the area of the region bounded by the polar curves $r = \sqrt{3} cos (θ)$ $r=sqrt(3)cos(theta)$ and $r = sin (θ)$ $r=sin(theta)$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the area of the region bounded by the polar curves r=√3cos(θ)r=sqrt(3)cos(theta) and r=sin(θ)r=sin(theta) ?

1 Answer

Related questions

Impact of this question

How do you find the area of the region bounded by the polar curves $r = \sqrt{3} cos (θ)$ $r=sqrt(3)cos(theta)$ and $r = sin (θ)$ $r=sin(theta)$ ?