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Find the centroid of the plane region shown below:

enter image description here

where the left and the right side is given by $x^2-y^2=4$.

Answer:

Its area is $$A =\int_{0}^{4} \int_{-\sqrt{4+y^2}}^{\sqrt{4+y^2}} dxdy \ =23.66$$

Thus if $ (\bar x, \bar y) \ $ be the centroid. How can I determine $ \bar x $ and $\bar y$ ?

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2 Answers 2

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Hint. Note that by symmetry $\bar x=0$ and $$\bar y=\frac{1}{A}\int_{y=0}^{4} \int_{x=-\sqrt{4+y^2}}^{\sqrt{4+y^2}} ydxdy=\frac{1}{A}\int_{0}^{4} \sqrt{4+y^2} (2y) dy.$$

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  • $\begingroup$ I am little confused whether the area of the $ beam $ $ \ x^2-y^2=4 \ $ is correct what I calculated . kindly give me idea how to find the area of the $beam $ with sides $ x^2-y^2=4 \ $ $\endgroup$
    – MAS
    Commented Nov 15, 2017 at 13:29
  • $\begingroup$ Your evaluation of $A$ is correct. $\endgroup$
    – Robert Z
    Commented Nov 15, 2017 at 13:34
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Divide the beam into three parts. The part with x going from -2 to 2 is a rectangle with width 4 and height 4 so its area is 16. By "symmetry" the two end regions, with x going from -4 to -2 and with x going from 2 to 4 are clearly the same so it is sufficient to find the area of the region with x going from 2 to 4. Take y going from 0 to 4. For each y, x goes from 0 to the graph $x^2- y^2= 4$ so $x^2= 4+ y^2$, $x= \sqrt{4+ y^2}$ (positive because x is between 2 and 4).

The area of that region is $A= \int_0^4 \sqrt{4+ y^2} dy$ and the cross section area of the beam is $16+ 2A$.

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