Find the centroid of the solid shown below:
Answers:
Since the solid is symmetric about x-axis, the y-coordinate of the centroid is $ 0 \ $ , i.e., $ \ \bar y=0 \ $
But how to find the other coordinates of the centroid $ \ (\bar x, \bar y, \bar z) \ $
In that case we have to find the volume of the volume closed by $ \ x=y^2 \ \ and \ \ x+z=1 \ $
But I am getting stuck to find the volume.
Help me out
$$\bar x = \frac1V \int_0^1 \int_{-\sqrt x}^{\sqrt x}\int_0^{1-x} x \mathrm d x\mathrm d y\mathrm d z = \frac1V\int_0^1 2x\sqrt x (1-x) \mathrm dx = 2\left[\frac{x^{\frac{5}{2}}}{\frac{5}{2}} - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right]_0^1 = \frac1V\frac{8}{35}$$ and $$\bar z = \frac1V\int_0^1 \int_{-\sqrt x}^{\sqrt x}\int_0^{1-x} z \mathrm d x\mathrm d y\mathrm d z = \frac1V\int_0^1 \sqrt x (1-x)^2 \mathrm dx = \frac1V \int_0^1 (\sqrt x -2x\sqrt x +x^2\sqrt x ) \mathrm d x = \ldots$$ and $$V = \int_0^1 \int_{-\sqrt x}^{\sqrt x}\int_0^{1-x} \mathrm d x\mathrm d y\mathrm d z = \int_0^1 2\sqrt x(1-x)\mathrm dx =\ldots $$