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Find the area enclosed between the circles:

$x^2 + (y- 3.5)^2 = (3.5)^2$
and
$(x-3.5)^2 + y^2 = (3.5)^2$

I tried using definite Integrals but I was unable to frame the equation. Please help me

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  • $\begingroup$ @imranfat thanks for the edit $\endgroup$ – David May 28 '18 at 15:51
  • $\begingroup$ Start by drawing a figure. You must find the intersections of the two circles $\endgroup$ – Andrei May 28 '18 at 15:54
  • $\begingroup$ @Andrei Found it at (3.5,3.5) , (0.0) what to do next? $\endgroup$ – David May 28 '18 at 15:55
  • $\begingroup$ @Andrei Can't proceed thereafter. Please help $\endgroup$ – David May 28 '18 at 15:55
  • $\begingroup$ @David Hello, Please look at the two answers provided. That ought to do it for you $\endgroup$ – imranfat May 28 '18 at 15:56
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HINT

Take a look to the graph

enter image description here

and use that for a quarter

  • $A_{circle}=\frac{\pi R^2}4$
  • $A_{triangle}=\frac12R^2$
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  • $\begingroup$ Thanks you but integrator provided me what i wanted $\endgroup$ – David May 28 '18 at 16:03
  • $\begingroup$ @David Did you get how to use the given hint? we do not need integral to calculate the area between the circles. $\endgroup$ – user May 28 '18 at 16:12
  • $\begingroup$ @please help me dealing with two different answers $\endgroup$ – David May 28 '18 at 16:16
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    $\begingroup$ Indeed for the second circle the function to be considered is $$x^2 + (y- 3.5)^2 = (3.5)^2\implies (y- 3.5)^2 = (3.5)^2-x^2 \implies (y- 3.5) =-\sqrt{ (3.5)^2-x^2} \implies y=3.5-\sqrt{ (3.5)^2-x^2} $$ $\endgroup$ – user May 28 '18 at 16:44
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    $\begingroup$ @TheIntegrator Yes indeed we need to take the lower part of the circle that is $$(y- 3.5) =-\sqrt{ (3.5)^2-x^2} $$ $\endgroup$ – user May 28 '18 at 16:46
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Sometimes graphing helps a lot. You can easily verify that the circles meet at the Origin and at $(3.5,3.5)$. You can also (with or without calculus) verify that at both points of intersection, the circles met at a right angle (Orthogonal intersection); that is, their respective tangents meet at a right angle. Now when you connect the points of intersection with a line segment, you "split up" the region they share into two equal regions. Now with some formulas of circle segment and circle sector, this is no longer a challenge

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  • $\begingroup$ It would be great if you show me the equation or steps $\endgroup$ – David May 28 '18 at 15:57
  • $\begingroup$ I got stucked taking limits from 0 to (7/2) and substracting $\endgroup$ – David May 28 '18 at 15:58
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    $\begingroup$ These formulas are provided by Gimusi. Essentially, you find the are of the circle sector (which is a quarter of the whole circle) and you subtract from it the area of the right triangle (base and height is the radius of the circle). Then double up. $\endgroup$ – imranfat May 28 '18 at 16:00
  • $\begingroup$ I already noted that one $\endgroup$ – David May 28 '18 at 16:07
  • $\begingroup$ @imranafat help me look at integrator comment section $\endgroup$ – David May 28 '18 at 16:16
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You need to find the points where they intersect, since you already fund that to be $(0,0)$ and $(3.5,3.5)$ I'll skip that step;

next to find the area you find the definite integral ;

$A = \displaystyle\int_0^{3.5}\bigg(\sqrt{(3.5)^2-(x-3.5)^2}+\sqrt{(3.5)^2-x^2}-3.5\bigg)\,dx$

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  • $\begingroup$ @Integrator But that doesn't help sir. I did the same. I got around 12. $\endgroup$ – David May 28 '18 at 16:00
  • $\begingroup$ Ok accepting your answer $\endgroup$ – David May 28 '18 at 16:02
  • $\begingroup$ @TheIntegrator It was the numerical value the OP was looking for, nothing else. $\endgroup$ – imranfat May 28 '18 at 16:04
  • $\begingroup$ @imranfat,i didnt realise that , sorry. $\endgroup$ – The Integrator May 28 '18 at 16:05
  • $\begingroup$ @David Id rather you just vote my answer but accept gimusi's answer as it better imo $\endgroup$ – The Integrator May 28 '18 at 16:06
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You can use simple geometry, like in the answers provided by other people. If you must use integrals, @gimusi provided a figure for you. Integrate $y_U(x)-y_L(x)$ for $x$ between $0$ and $3.5$. $y_U$ is the equation of the upper curve, $y_l$ is the equation of the lower curve.

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  • $\begingroup$ Integrator gave the answer $\endgroup$ – David May 28 '18 at 16:03
  • $\begingroup$ help me look at integrator comment section $\endgroup$ – David May 28 '18 at 16:16

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