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June 29th, 2018, 08:47 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 10:55 PM.
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June 29th, 2018, 09:23 PM   #2
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Math Focus: Calculus/ODEs
I would express the area $\displaystyle A$ as:

$\displaystyle A=2\left(\int_{0}^{1} \sqrt{3x}\,dx+\int_{1}^{2} \sqrt{4-x^2}\,dx\right)$

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June 29th, 2018, 11:18 PM   #3
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Was this a multiple choice question? Also, was it given as a calculus problem?
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June 29th, 2018, 11:29 PM   #4
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Quote:
Originally Posted by skipjack View Post
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 02:07 AM.
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June 29th, 2018, 11:49 PM   #5
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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:

$\displaystyle A=\frac{4}{3}(\sqrt{3})(1)+ \frac{2^2}{2}\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}+4\pi}{3}$
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June 30th, 2018, 11:52 AM   #6
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$\displaystyle x^2+ y^2= 4$ 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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