Answer: Either

The mean value of a function on \(0\le x\le 3\) is \(\frac{1}{3}\int_0^3 y\,dx\). Here \(y=\frac{1}{3}x^{1/2}(3-x)\), so

\(\displaystyle \bar y=\frac{1}{3}\int_0^3 \frac{1}{3}x^{1/2}(3-x)\,dx=\frac{4\sqrt3}{15}\).

Also, \(\frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right)\), so the arc length is

\(\displaystyle s=\int_0^3 \frac12\left(x^{-1/2}+x^{1/2}\right)\,dx=2\sqrt3\).

The surface area generated when \(C\) is rotated about the \(x\)-axis is

\(\displaystyle S=2\pi\int_0^3 y\frac{ds}{dx}\,dx=3\pi\).

Or

The eigenvalues of \(A\) are \(1,2,3\), with corresponding eigenvectors that may be taken as \(\begin{pmatrix}0\\-2\\1\end{pmatrix}\), \(\begin{pmatrix}1\\1\\0\end{pmatrix}\), and \(\begin{pmatrix}2\\2\\1\end{pmatrix}\).

The planes with the required property are

\(x-y-2z=0\), \(2x-y-2z=0\), and \(x-y=0\).

Either

For a curve given by \(y=f(x)\) on \(0\le x\le 3\), the mean value of \(y\) is

\(\displaystyle \bar y=\frac{1}{3-0}\int_0^3 y\,dx.\)

Here

\(\displaystyle y=\frac13 x^{1/2}(3-x)=x^{1/2}-\frac13 x^{3/2}.\)

So

\(\displaystyle \bar y=\frac13\int_0^3 \left(x^{1/2}-\frac13 x^{3/2}\right)dx.\)

Integrating,

\(\displaystyle \int \left(x^{1/2}-\frac13 x^{3/2}\right)dx=\frac23 x^{3/2}-\frac{2}{15}x^{5/2}.\)

Hence

\(\displaystyle \bar y=\frac13\left[\frac23 x^{3/2}-\frac{2}{15}x^{5/2}\right]_0^3.\)

Now \(3^{3/2}=3\sqrt3\) and \(3^{5/2}=9\sqrt3\), so

\(\displaystyle \bar y=\frac13\left(\frac23\cdot 3\sqrt3-\frac{2}{15}\cdot 9\sqrt3\right)=\frac{4\sqrt3}{15}.\)

Next, to find \(\frac{ds}{dx}\), use

\(\displaystyle \frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}.\)

Differentiating \(y=x^{1/2}-\frac13x^{3/2}\),

\(\displaystyle \frac{dy}{dx}=\frac12 x^{-1/2}-\frac12 x^{1/2}.\)

Therefore

\(\displaystyle \left(\frac{dy}{dx}\right)^2=\frac14\left(x^{-1}-2+x\right),\)

and so

\(\displaystyle \frac{ds}{dx}=\sqrt{1+\frac14\left(x^{-1}-2+x\right)}=\sqrt{\frac14\left(x^{-1}+2+x\right)}.\)

This simplifies to

\(\displaystyle \frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right).\)

The arc length is then

\(\displaystyle s=\int_0^3 \frac12\left(x^{-1/2}+x^{1/2}\right)dx.\)

So

\(\displaystyle s=\frac12\left[2x^{1/2}+\frac23x^{3/2}\right]_0^3=\sqrt3+\sqrt3=2\sqrt3.\)

For the surface area generated by rotation about the \(x\)-axis, use

\(\displaystyle S=2\pi\int_0^3 y\frac{ds}{dx}\,dx.\)

Substitute \(y=x^{1/2}-\frac13x^{3/2}\) and \(\frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right)\):

\(\displaystyle S=2\pi\int_0^3 \left(x^{1/2}-\frac13x^{3/2}\right)\frac12\left(x^{-1/2}+x^{1/2}\right)dx.\)

Simplifying the integrand,

\(\displaystyle \left(x^{1/2}-\frac13x^{3/2}\right)\frac12\left(x^{-1/2}+x^{1/2}\right)=\frac12\left(1+\frac23x-\frac13x^2\right).\)

Hence

\(\displaystyle S=\pi\int_0^3 \left(1+\frac23x-\frac13x^2\right)dx.\)

Integrating,

\(\displaystyle S=\pi\left[x+\frac13x^2-\frac19x^3\right]_0^3=\pi(3+3-3)=3\pi.\)

Or

The matrix is

\(\displaystyle A=\begin{pmatrix}1&1&2\\0&2&2\\-1&1&3\end{pmatrix}.\)

Its characteristic equation is found from \(\det(A-\lambda I)=0\):

\(\displaystyle \det\begin{pmatrix}1-\lambda&1&2\\0&2-\lambda&2\\-1&1&3-\lambda\end{pmatrix}=0.\)

This gives

\(\displaystyle \lambda^3-6\lambda^2+11\lambda-6=0,\)

which factors as

\(\displaystyle (\lambda-1)(\lambda-2)(\lambda-3)=0.\)

So the eigenvalues are \(\lambda=1,2,3\).

For \(\lambda=1\), solve \((A-I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}0&1&2\\0&1&2\\-1&1&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0,\)

which gives \(y=-2z\), \(x=z\). Taking \(z=1\), one eigenvector is

\(\displaystyle \mathbf{e}_1=\begin{pmatrix}0\\-2\\1\end{pmatrix}.\)

For \(\lambda=2\), solve \((A-2I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}-1&1&2\\0&0&2\\-1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0.\)

From the second row, \(z=0\), and then \(x=y\). Taking \(x=1\), an eigenvector is

\(\displaystyle \mathbf{e}_2=\begin{pmatrix}1\\1\\0\end{pmatrix}.\)

For \(\lambda=3\), solve \((A-3I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}-2&1&2\\0&-1&2\\-1&1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0.\)

From the second row, \(y=2z\). Then the third row gives \(x=2z\). Taking \(z=1\), an eigenvector is

\(\displaystyle \mathbf{e}_3=\begin{pmatrix}2\\2\\1\end{pmatrix}.\)

Now let \(\mathbf{e}\) and \(\mathbf{f}\) be two linearly independent eigenvectors with eigenvalues \(\lambda\) and \(\mu\), and let \(\Pi\) be the plane through the origin containing them. Any point \(\mathbf{r}\) in \(\Pi\) can be written as

\(\displaystyle \mathbf{r}=a\mathbf{e}+b\mathbf{f}\)

for some scalars \(a,b\). Then

\(\displaystyle T(\mathbf{r})=A\mathbf{r}=A(a\mathbf{e}+b\mathbf{f})=aA\mathbf{e}+bA\mathbf{f}=a\lambda\mathbf{e}+b\mu\mathbf{f}.\)

This is again a linear combination of \(\mathbf{e}\) and \(\mathbf{f}\), so \(T(\mathbf{r})\) also lies in \(\Pi\). Therefore all points of \(\Pi\) are mapped onto points of \(\Pi\).

The three planes generated by pairs of eigenvectors are obtained by taking normal vectors as cross products:

\(\displaystyle \mathbf{e}_1\times\mathbf{e}_2=\begin{pmatrix}0\\-2\\1\end{pmatrix}\times\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}-1\\1\\2\end{pmatrix},\)

so one plane is \(x-y-2z=0\).

\(\displaystyle \mathbf{e}_1\times\mathbf{e}_3=\begin{pmatrix}0\\-2\\1\end{pmatrix}\times\begin{pmatrix}2\\2\\1\end{pmatrix}=\begin{pmatrix}2\\-1\\-2\end{pmatrix},\)

so another plane is \(2x-y-2z=0\).

\(\displaystyle \mathbf{e}_2\times\mathbf{e}_3=\begin{pmatrix}1\\1\\0\end{pmatrix}\times\begin{pmatrix}2\\2\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\0\end{pmatrix},\)

so the third plane is \(x-y=0\).