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I'm trying to solve the following problem:

Find the area of the circle $x^2+y^2=8$, which is interior to the parabola $y^2=2x$.

I have my own solution and I want to verify whether it is correct. My solution is to take the $1/4$ of the area of the circle, which is $\int_{0}^{\sqrt{8}}\sqrt{-x^2+8}$ and subtract the rest of the area we don't want, which is : $\int_{0}^{2}\sqrt{-x^2+8} -\sqrt{2x}$ (it is the 1/4 of the area of the circle minus the area under the parabole).

Therefore the area is $2*(\int_{0}^{\sqrt{8}}(\sqrt{-x^2+8} )- \int_{0}^{2}(\sqrt{-x^2+8} -\sqrt{2x}))$.

Is my solution correct?

Thanks

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  • $\begingroup$ Yes the solution is correct. But you can avoid calculating the integral that gives the area of the quarter disc with center 0 and radius $\sqrt{8}$ ($\pi (\sqrt{8})^2/4$) $\endgroup$
    – Jean-Claude Colette
    Commented Jan 18, 2020 at 14:12
  • $\begingroup$ Please note that these integral expressions are not legitimate without the differential $dx$. $\endgroup$
    – electronpusher
    Commented Jan 20, 2020 at 9:51

2 Answers 2

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Take a look at the picture below:

enter image description here

The area that you're after is the area of a quarter of the given circle (in this case, that's $2\pi$) plus twice the area of the region below the parabola and above the line $y=x$.

So, the area is equal to$$2\pi+2\int_0^2\sqrt{2x}-x\,\mathrm dx=2\pi+\frac43.$$

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  • $\begingroup$ I'm calculating the following area, right? i.imgur.com/WiFi7cJ.png (sketched it in GeoGebra). Is my solution calculating the area I sketched? Thanks $\endgroup$
    – LukasT
    Commented Jan 18, 2020 at 14:11
  • $\begingroup$ Yes, that is the area that I computed too. But I was unable to understand your computations. $\endgroup$
    – José Carlos Santos
    Commented Jan 18, 2020 at 14:22
  • $\begingroup$ Could you please explain why do you add twice the area below the parabola and above the line y=x? That step is really not clear to me. $\endgroup$
    – LukasT
    Commented Jan 18, 2020 at 14:36
  • $\begingroup$ Because that regian appears twice, once in the upper half of the picture and once below. $\endgroup$
    – José Carlos Santos
    Commented Jan 18, 2020 at 14:40
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Looking at it graphically we can notice that this area is equal to the area of a semicircle minus the area under the quadratic: $$A=\frac{\pi}{2}r^2-\int_{-r}^rxdy=\frac{\pi}{2}r^2-\int_{-r}^r\frac{y^2}{2}dy$$ Now just work this out and sub in $r=2\sqrt{2}$

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  • $\begingroup$ But is my solution correct? $\endgroup$
    – LukasT
    Commented Jan 18, 2020 at 14:04
  • $\begingroup$ That gives the other area, from the question it is unclear which one you are trying to calculate $\endgroup$
    – Henry Lee
    Commented Jan 18, 2020 at 14:07
  • $\begingroup$ @ I am trying to calculate the following area: i.imgur.com/WiFi7cJ.png . $\endgroup$
    – LukasT
    Commented Jan 18, 2020 at 14:10
  • $\begingroup$ Yep from your diagram I believe that your solution calculates the little near-triangles on either side of this area $\endgroup$
    – Henry Lee
    Commented Jan 18, 2020 at 14:13
  • $\begingroup$ But why is that? The quarter of the circle is $\int_{0}^{\sqrt{8}}\sqrt{-x^2+8}$, right? And now if I subtract from the quarter the upper triangle, the area of which should be $\int_{0}^{2}\sqrt{-x^2+8} -\sqrt{2x}$, I should get 1/2 of the area I want, right? Not really sure where I did the mistake. $\endgroup$
    – LukasT
    Commented Jan 18, 2020 at 14:17

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