January 4th, 2018, 12:10 PM  #1 
Member Joined: Oct 2012 Posts: 70 Thanks: 0  Find area
Evaluate area of the region enclosed by the circle x^2+y^2=4 that which is above the line y=1 and below the line y=sqrt(3)x

January 4th, 2018, 12:54 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Have you drawn a graph? $x^2+ y^2= 4$ is, as you say, a circle, with center at (0, 0) and radius 2. $y= \sqrt{3}x$ is a straight line. That line intersects y= 1 where $\sqrt{3}x= 1$ so $x= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}$. It intersects the circle where $x^2+ 3x^2= 4x^2= 4$ so x= 1. The region inside that circle, above y= 1 and under $y= \sqrt{3}x$ can be broken into two parts. In the first, for $0\le x\le \frac{\sqrt{3}}{3}$, y ranges from 1 to $\sqrt{3}x$. The area of that region is $\sqrt{3}\int_0^{\frac{\sqrt{3}}{3}} x 1 dx$ which is, of course, the area of the triangle with base $\frac{\sqrt{3}}{3}$ and height $\sqrt{3} 1$. The area of the second region, from $x= \frac{\sqrt{3}}{3}$ to $x= \sqrt{3}$, y from 1 to $y= \sqrt{4 x^2}$, is $\int_{\frac{\sqrt{3}}{3}}^\sqrt{3} \sqrt{4 x^2}dx$. The entire area is the sum or those two. 
January 4th, 2018, 02:26 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,781 Thanks: 1432 
$\displaystyle A = \int_{\pi/6}^{\pi/3} 2  \dfrac{\csc^2{\theta}}{2} \, d\theta$


Tags 
area, find 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
How do I find the area?  Apple11  Calculus  1  February 21st, 2017 06:17 PM 
How to find area ?  MMath  Calculus  3  June 21st, 2016 05:46 AM 
find Area of ABC  brhum  Geometry  11  June 9th, 2014 06:07 AM 
find the area help?  johnny993  Algebra  1  November 12th, 2013 10:17 PM 
Find area  450081592  Calculus  7  January 19th, 2010 03:45 AM 