June 29th, 2018, 07:47 PM  #1 
Member Joined: Aug 2017 From: India Posts: 45 Thanks: 2  Finding Area
The problem is and only things I know is one is circle of radius 2 and the other is parabola, but not sure how to proceed to find area. Please advise.
Last edited by skipjack; June 29th, 2018 at 09:55 PM. 
June 29th, 2018, 08:23 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 
I would express the area $\displaystyle A$ as: $\displaystyle A=2\left(\int_{0}^{1} \sqrt{3x}\,dx+\int_{1}^{2} \sqrt{4x^2}\,dx\right)$ Can you proceed? 
June 29th, 2018, 10:18 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,528 Thanks: 1750 
Was this a multiple choice question? Also, was it given as a calculus problem?

June 29th, 2018, 10:29 PM  #4  
Member Joined: Aug 2017 From: India Posts: 45 Thanks: 2  Quote:
$\displaystyle 1/(2\sqrt{3}) + \pi/3, 1/\sqrt{3} + 2\pi/3, 1/(2\sqrt{3}) + 2\pi/3, 1/\sqrt{3} + 4\pi/3 $ Last edited by skipjack; June 30th, 2018 at 01:07 AM.  
June 29th, 2018, 10:49 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 
I would use the formula for a parabolic cap (which I derived using calculus), along with the difference between the area of an isosceles triangle and circular sector to write: $\displaystyle A=\frac{4}{3}(\sqrt{3})(1)+ \frac{2^2}{2}\left(\frac{2\pi}{3}\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}+4\pi}{3}$ 
June 30th, 2018, 10:52 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
$\displaystyle x^2+ y^2= 4$ is, of course, the circle with center at the origin and radius 2. $\displaystyle y^2= 3x$ is a parabola with vertex at the origin and horizontal axis, The two graphs cross where $\displaystyle x^2+ y^2= x^2+ 3x= 4$ so $\displaystyle x^2+ 3x 4= (x 1)(x+ 4)= 0$. Since $\displaystyle 3x= y^2$ cannot be negative, x= 1. That is why MarkFL has his first integral from x= 0 to x= 1 and his second from x= 1 to x= 2.


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