Answer: Since the region lies under the curve \(y=\lambda x^2\) from \(x=0\) to \(x=a\), its area is
\(A=\int_0^a \lambda x^2\,dx=\lambda \left[\frac{x^3}{3}\right]_0^a=\frac{\lambda a^3}{3}.\)
The centroid formula for the \(y\)-coordinate of a region under \(y=f(x)\) is
\(\bar y=\frac{1}{A}\int_0^a \frac{1}{2}y^2\,dx.\)
So here
\(\bar y=\frac{1}{\lambda a^3/3}\int_0^a \frac{1}{2}(\lambda x^2)^2\,dx=\frac{3}{\lambda a^3}\cdot \frac{1}{2}\lambda^2\int_0^a x^4\,dx.\)
Now
\(\int_0^a x^4\,dx=\left[\frac{x^5}{5}\right]_0^a=\frac{a^5}{5},\)
hence
\(\bar y=\frac{3}{\lambda a^3}\cdot \frac{1}{2}\lambda^2\cdot \frac{a^5}{5}=\frac{3\lambda a^2}{10}.\)
Given that the \(y\)-coordinate of the centroid is \(a\), we have
\(a=\frac{3\lambda a^2}{10}.\)
Since \(a\gt 0\), divide by \(a\):
\(1=\frac{3\lambda a}{10}\).
Therefore
\(\lambda=\frac{10}{3a}.\)
Let the region be the area under \(y=\lambda x^2\) from \(x=0\) to \(x=a\).
Its area is
\(A=\int_0^a y\,dx=\int_0^a \lambda x^2\,dx=\lambda \left[\frac{x^3}{3}\right]_0^a=\frac{\lambda a^3}{3}.\)
For a region under a curve, the \(y\)-coordinate of the centroid is
\(\bar y=\frac{1}{A}\int_0^a \frac{1}{2}y^2\,dx.\)
So
\(\bar y=\frac{1}{\lambda a^3/3}\int_0^a \frac{1}{2}(\lambda x^2)^2\,dx.\)
Now
\(\int_0^a \frac{1}{2}(\lambda x^2)^2\,dx=\frac{\lambda^2}{2}\int_0^a x^4\,dx=\frac{\lambda^2}{2}\left[\frac{x^5}{5}\right]_0^a=\frac{\lambda^2 a^5}{10}.\)
Hence
\(\bar y=\frac{\lambda^2 a^5/10}{\lambda a^3/3}=\frac{3\lambda a^2}{10}.\)
Given that the centroid has \(y\)-coordinate \(a\),
\(a=\frac{3\lambda a^2}{10}.\)
Since \(a\gt 0\), divide by \(a\):
\(1=\frac{3\lambda a}{10}.\)
Therefore
\(\lambda=\frac{10}{3a}.\)
