Find the centroid of the homogeneous figure bounded
Find the centroid of the homogeneous figure bounded by the curves $y=2x^3$ and $y^2=2x$.
Explanation of Method
Finding the two intersection points of the cubic and the parabola gives the bounds of integration; the area and static moments then follow from ordinary single-variable integrals of the vertical strip between the two curves.
- Find the intersection points.
- Compute the area between the curves.
- Compute the static moments $M_y,M_x$.
- Divide to get the centroid.
Step 1 · Intersection points
Substituting $y=2x^3$ into $y^2=2x$: $4x^6=2x\implies2x(2x^5-1)=0$, so $x=0$ or $x=2^{-1/5}$. The curves meet at $(0,0)$ and $\left(2^{-1/5},2^{2/5}\right)$. On $0\le x\le2^{-1/5}$, the parabola $y=\sqrt{2x}$ lies above the cubic $y=2x^3$.
Step 2 · Area
$$ A = \int_0^{2^{-1/5}}\left(\sqrt{2x}-2x^3\right)dx = \left[\frac{2\sqrt2}3x^{3/2}-\frac{x^4}2\right]_0^{2^{-1/5}} = \frac56\cdot2^{-4/5}. $$
Step 3 · Static moments
$$ M_y = \int_0^{2^{-1/5}}x\left(\sqrt{2x}-2x^3\right)dx = \left[\frac{2\sqrt2}5x^{5/2}-\frac{2x^5}5\right]_0^{2^{-1/5}} = \frac15, $$
$$ M_x = \frac12\int_0^{2^{-1/5}}\left[(2x)-(2x^3)^2\right]dx = \frac12\left[x^2-\frac{4x^7}7\right]_0^{2^{-1/5}} = \frac5{14}\cdot2^{-2/5}. $$
Step 4 · Centroid
$$ \bar x = \frac{M_y}A = \frac{1/5}{\frac56\cdot2^{-4/5}} = \frac{6\cdot2^{4/5}}{25} = \frac{6\sqrt[5]{16}}{25}, $$
$$ \bar y = \frac{M_x}A = \frac{\frac5{14}\cdot2^{-2/5}}{\frac56\cdot2^{-4/5}} = \frac37\cdot2^{2/5} = \frac{3\sqrt[5]4}7. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: $A\approx0.478624$, $\bar x\approx0.417864$, $\bar y\approx0.565503$, matching $\dfrac56\cdot2^{-4/5},\,\dfrac{6\sqrt[5]{16}}{25},\,\dfrac{3\sqrt[5]4}7$ exactly.
Result
The centroid of the region is: