Given this solid in the picture, calculate the centroid: enter image description here

First things first: I calculated the filled area: $$A = \int_{0}^af(x_1)dx = b\int_{0}^ax^{\frac{1}{3}}dx = b\cdot\frac{3}{4}\cdot\sqrt[3]{a^4}$$ I calculated the $\overline{x}_{1}$ coordinate: $$\overline{x}_{1} = \dfrac{1}{A}\cdot\int_{0}^a x_1da=\dots=\dfrac{4}{7}a.$$

And this does match the given solution but I do not understand how one could calculate the $\overline{x}_2$ coordinate of the centroid. I am having trouble because the centroid does not contact the $x_2$ axis.

First I calculated the inverse of $f(x_1)$ to get $\overline{x}_2$. And then I was stuck. How do I need to go on from here?


Hint. In order to find $x_1$, we evaluate the average value of the coordinate $x_1$ over the yellow region $R$, that is $$\overline{x}_{1} = \dfrac{1}{|R|}\iint_R x_1dx_1dx_2=\dfrac{1}{|R|}\int_{x_1=0}^a x_1\left(\int_{x_2=0}^{f(x_1)}dx_2\right)dx_1=\dfrac{1}{|R|}\int_{x_1=0}^a x_1f(x_1)dx_1$$ where $|R|$ is the area of the region $R$. Similarly, for $x_2$, we evaluate the integral $$\overline{x}_{2} = \dfrac{1}{|R|}\iint_R x_2dx_1dx_2=\dfrac{1}{|R|}\int_{x_1=0}^a \left(\int_{x_2=0}^{f(x_1)}x_2dx_2\right)dx_1.$$ Can you take it from here?

  • $\begingroup$ I will try. Thank you :) $\endgroup$ – Finn Eggers Nov 21 '17 at 15:01

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