The centroid for the given parabola from the x-axis is:

This question was previously asked in

PGCIL DT Civil 2018 Official Paper

Option 4 : 3h/10

**Concept:**

**The general formula for centroid is given by:**

\(\bar{x}=\frac{∫ xdA}{∫ dA}\) & \(\bar{y}=\frac{∫ ydA}{∫ dA}\)

For a parabola of y = kx^{2} as given in the figure below:

Area of the strip is (h - y)dx ⇒ (h - kx^{2})dx

where \(k=\frac{h}{a^2}\)

**Calculation:**

**Given:**

Consider that the vertex is at origin (0, 0)

The given parabola is y = -kx2

\(A=\int dA =\int_0^a2(h\;-\;y)dx=\int_0^a2(h\;+\;kx^2)dx\)

\(A=2h[x]_0^a\;+\;\frac{k}{3}[x^3]_0^a\)

\(A=2ha\;+\;\frac{ka^3}{3}\)

Putting \(k=\frac{h}{a^2}\)

\(A=2ha\;+\;\frac{h}{a^2}\times\frac{a^3}{3}=\frac{8ha}{3}\)

\(\bar{y}=\frac{∫ ydA}{∫ dA}\)

\(\int ydA=2\int\left(\frac{h+y}{2}\right)(h\;+\;kx^2)dx\)

\(\int ydA=\int\left({h\;-\;kx^2}\right)(h\;+\;kx^2)dx\)

\(\int ydA=\int\left({h^2\;-\;k^2x^4}\right)dx\)

\(\int ydA=h^2[x]_0^a\;-\;\frac{k^2}{5}[x^5]_0^a\)

\(\int ydA=h^2a\;-\;\frac{h^2}{5a^4}[a^5]\Rightarrow \frac{4h^2a}{5}\)

\(\bar{y}=\frac{∫ ydA}{A}=\frac{\frac{4h^2a}{5}}{\frac{3}{8ha}}=\frac{3h}{10}\)