Polar decomposition of a deformation gradient for the simple shear Consider the following deformation $\boldsymbol{x} = \boldsymbol{f}(\boldsymbol{p})$ definied by
$$
\begin{cases}
x_1 = p_1 + \gamma p_2 \\
x_2 = p_2 \\
x_3 = p_3
\end{cases}
$$
which corresponds to the simple shear.
it's immediate to find the deformation gradient $F$ to be
$$\begin{bmatrix}
1 & \gamma & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
This tensor (it's a matrix) have positive determinant, so we can compute the Polar decomposition, i.e. $F=VR=RU$ where $R$ is a rotation and $U,V$ are symmetric and positive definite.
My goal is to compute $U$, which is known to be $\sqrt{F^T F}$, which is not easy at all. Is there any way to compute it, without computing the square root of the matrix $ F^T F$?
 A: Instead of calculating $\sqrt{F^TF}$, you can construct a U and R with some variables and solve for them individually.
Let's say,
$$
    \boldsymbol{U}=\quad
    \begin{pmatrix}
    a & b & 0 \\
    b & c & 0 \\
    0 & 0 & 1
    \end{pmatrix}
    \quad
    \boldsymbol{R}=\quad
    \begin{pmatrix}
    x & y & 0 \\
    -y & x & 0 \\
    0 & 0 & 1
    \end{pmatrix}
    \quad
$$
Here, U(1,2) = U(2,1), because it is symmetric. Similarly for R, because it is anti-symmetric.
Using, $F=R.U$, we get,
$$
ax-by=1\\
    bx-cy=\gamma\\
    bx+ay=0\\
    cx+by=1\\
    x^2+y^2=1
$$
Solving the above equations, we get (I used Mathematica),
$$
\boldsymbol{U}=\quad
        \begin{pmatrix}
        \frac{2}{\sqrt{4+\gamma^2}} & \frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\
        \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}}+\frac{\gamma^2}{\sqrt{4+\gamma^2}} & 0 \\
        0 & 0 & 1
        \end{pmatrix}
        \quad\\
    \boldsymbol{R}=\quad
        \begin{pmatrix}
        \frac{2}{\sqrt{4+\gamma^2}} & -\frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\
        \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}} & 0 \\
        0 & 0 & 1
        \end{pmatrix}
        \quad
$$
