# Find the centroid of the area of of a plane region with sides given by $x^2-y^2=4$

Find the centroid of the plane region shown below:

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

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 $\bar y$ ?

Hint. Note that by symmetry $\bar x=0$ and $$\bar y=\frac{1}{A}\int_{y=0}^{4} \int_{x=-\sqrt{4+y^2}}^{\sqrt{4+y^2}} ydxdy=\frac{1}{A}\int_{0}^{4} \sqrt{4+y^2} (2y) dy.$$
• 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 \$
• Your evaluation of $A$ is correct. Commented Nov 15, 2017 at 13:34
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 $16+ 2A$.