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$?
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), $$ \boldsymbol{U}=\quad \begin{pmatrix} \frac{2}{\sqrt{4+\gamma^2}} & \frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\ \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}}+\frac{\gamma^2}{\sqrt{4+\gamma^2}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \quad\\ \boldsymbol{R}=\quad \begin{pmatrix} \frac{2}{\sqrt{4+\gamma^2}} & -\frac{\gamma}{\sqrt{4+\gamma^2}} & 0 \\ \frac{\gamma}{\sqrt{4+\gamma^2}} & \frac{2}{\sqrt{4+\gamma^2}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \quad $$