Find the centroid of the plane figure bounded by ay²=x³
Find the centroid of the plane figure bounded by $ay^2=x^3$ and the line $x=a$ (where $a>0$).
Explanation of Method
Write the upper branch as $y=x^{3/2}/\sqrt a$, note $M_x=0$ by symmetry about the $x$-axis, and compute the area and $M_y$ by direct power-rule integration.
Step 1: Area
By symmetry about the $x$-axis: \[A=2\int_0^a\frac{x^{3/2}}{\sqrt a}\,dx = \frac2{\sqrt a}\cdot\frac{2a^{5/2}}5=\frac{4a^2}5\]
Step 2: Moment about the $y$-axis
\[M_y=2\int_0^ax\cdot\frac{x^{3/2}}{\sqrt a}\,dx = \frac2{\sqrt a}\cdot\frac{2a^{7/2}}7=\frac{4a^3}7\]
Step 3: Moment about the $x$-axis
By symmetry, $M_x=0$.
Step 4: Centroid
\[\bar x=\frac{M_y}A=\frac{4a^3/7}{4a^2/5}=\frac{5a}7,\qquad \bar y=0\]
\[\boxed{\left(\frac{5a}7,0\right)\approx(0.714a,0)}\]
Step 5: Interpretation
The centroid sits well past the midpoint $x=a/2$, reflecting that the region widens rapidly (as $x^{3/2}$) toward $x=a$, concentrating more area there.
Result
Final answer:
\[\boxed{\left(\dfrac{5a}{7},\ 0\right) \approx (0.714a,\ 0)}\]