My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree1Thanks
  • 1 Post By skeeter
LinkBack Thread Tools Display Modes
January 4th, 2018, 11:10 AM   #1
Joined: Oct 2012

Posts: 77
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
fahad nasir is offline  
January 4th, 2018, 11:54 AM   #2
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

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.
Country Boy is offline  
January 4th, 2018, 01:26 PM   #3
Math Team
skeeter's Avatar
Joined: Jul 2011
From: Texas

Posts: 2,949
Thanks: 1555

$\displaystyle A = \int_{\pi/6}^{\pi/3} 2 - \dfrac{\csc^2{\theta}}{2} \, d\theta$
Attached Images
File Type: jpg Area_partial_sector.jpg (32.6 KB, 1 views)
Thanks from Country Boy
skeeter is offline  

  My Math Forum > College Math Forum > Calculus

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 05:17 PM
How to find area ? MMath Calculus 3 June 21st, 2016 04:46 AM
find Area of ABC brhum Geometry 11 June 9th, 2014 05:07 AM
find the area help? johnny993 Algebra 1 November 12th, 2013 09:17 PM
Find area 450081592 Calculus 7 January 19th, 2010 02:45 AM

Copyright © 2019 My Math Forum. All rights reserved.