Question

CentroidsFind the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.

Accepted Answer

Recall that the centroid (or center of mass) of a curve defined parametrically by $$x = x(t)$$ and $$y = y(t)$$ over an interval $$a \leq t \leq b is $$given by the coordinates $$\left( \bar{x}, \bar{y} \right)$$, where: Write the formulas for the centroid coordinates of a curve:  
$$\displaystyle \bar{x} = \frac{\int_a^b x(t) \sqrt{\left( \frac{dx}{dt} \$$right)^2$$ + \left( \frac{dy}{dt} \$$right)^2$$} \, dt}{\int_a^b \sqrt{\left( \frac{dx}{dt} \$$right)^2$$ + \left( \frac{dy}{dt} \$$right)^2$$} \, dt}$$  
and  
$$\displaystyle \bar{y} = \frac{\int_a^b y(t) \sqrt{\left( \frac{dx}{dt} \$$right)^2$$ + \left( \frac{dy}{dt} \$$right)^2$$} \, dt}{\int_a^b \sqrt{\left( \frac{dx}{dt} \$$right)^2$$ + \left( \frac{dy}{dt} \$$right)^2$$} \, dt}$$ Calculate the derivatives $$\frac{dx}{dt}$$ and $$\frac{dy}{dt}$$ for the given parametric equations:  
Since $$x = \cos t$$, then $$\frac{dx}{dt} = -\sin t.  
$$Since $$y = t + \sin t$$, then $$\frac{dy}{dt} = 1 + \cos t. $$Substitute $$x(t)$$, $$y(t)$$, $$\frac{dx}{dt}$$, and $$\frac{dy}{dt}$$ into the centroid formulas and set up the integrals over the interval $$0 \leq t \leq \pi$$:  
$$\displaystyle \bar{x} = \frac{\$$int_0$$^{\pi} \cos t \sqrt{(-\sin t)^2$ + (1 + \cos t)^2$} \, dt}{\$$int_0$$^{\pi} \sqrt{(-\sin t)^2$ + (1 + \cos t)^2$} \, dt}$$  
$$\displaystyle \bar{y} = \frac{\$$int_0$$^{\pi} (t + \sin t) \sqrt{(-\sin t)^2$ + (1 + \cos t)^2$} \, dt}{\$$int_0$$^{\pi} \sqrt{(-\sin t)^2$ + (1 + \cos t)^2$} \, dt}$$ Evaluate the integrals in the numerator and denominator separately (using appropriate integration techniques or numerical methods if necessary), then divide to find the coordinates $$\bar{x}$$ and $$\bar{y} of $$the centroid.

Centroids

Find the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.

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