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June 29th, 2018, 07:47 PM   #1
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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.
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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: 12,184 Thanks: 481 Math Focus: Calculus/ODEs I would express the area $\displaystyle A$ as: $\displaystyle A=2\left(\int_{0}^{1} \sqrt{3x}\,dx+\int_{1}^{2} \sqrt{4-x^2}\,dx\right)$ Can you proceed? Thanks from Country Boy and MathsLearner123
 June 29th, 2018, 10:18 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,191 Thanks: 1649 Was this a multiple choice question? Also, was it given as a calculus problem?
June 29th, 2018, 10:29 PM   #4
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Quote:
 Originally Posted by skipjack Was this a multiple choice question? Also, was it given as a calculus problem?
Yes, it is a multiple choice, but I can't confirm it is under calculus as it consists of all questions.
$\displaystyle 1/(2\sqrt{3}) + \pi/3, 1/\sqrt{3} + 2\pi/3, 1/(2\sqrt{3}) + 2\pi/3, 1/\sqrt{3} + 4\pi/3$

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: 12,184 Thanks: 481 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: 3,248 Thanks: 887 $\displaystyle x^2+ y^2= 4$ 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. Thanks from MarkFL and MathsLearner123

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