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Show that the area lying in the region $x\geq 4$ and between the circle $x^{2}+y^{2}=32$ , the $x$-axis and the tangent drawn at the point $(4,4)$ on the circle is $4\left( 4-\pi \right)$ square units.

I tried plotting the concerned region, which I think is the area highlighted in green in the figure below. But I did not get the answer as $4\left( 4-\pi \right)$ square units.

enter image description here

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    $\begingroup$ That is because I think the area in question falls outside the circle. It is the area outside the circle, above the x-axis and under the line $\endgroup$ – imranfat May 25 '18 at 17:42
  • $\begingroup$ @imranfat , do you mean to say that the area in question is the region $EHFE$? $\endgroup$ – rikodou sennin May 25 '18 at 17:44
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    $\begingroup$ Yes, that is what I think and the reason is because of the word "between" though I do admit, it is phrased a bit ambiguous $\endgroup$ – imranfat May 25 '18 at 17:44
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    $\begingroup$ It should be $EHFE$. I know you tagged it as integration, but to check its area, you can use geometry and have Area of EDF - Area of EDH. It is equal to $4(4-\pi)$ $\endgroup$ – Wyllich May 25 '18 at 17:52
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    $\begingroup$ @rikodou I believe that's what imranfat is saying. $AEHG$ is $8\pi$. $AEFG$ is $32$. $EHGF$ is the difference, $8(4-pi)$. Half of this is $EHF$ and your desired area! $\endgroup$ – Captain Morgan May 25 '18 at 17:55
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AHE is 1/8 of the circle, ADE is a right triangle, so DHE is their difference.

EDF is the same as EDA and DHE is the same as DHG.

This should get you whatever you want.

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