Step 1: Geometric interpretation (Steinmetz solid).
The region is the intersection of two perpendicular cylinders:
• \(x^2+y^2 \le 4\) (cylinder along the \(z\)-axis)
• \(x^2+z^2 \le 4\) (cylinder along the \(y\)-axis)
Their intersection is a classical Steinmetz solid.
Step 2: Use slicing with respect to \(x\).
Fix a value of \(x\).
From the first cylinder:
\[
y^2 \le 4-x^2
\]
From the second cylinder:
\[
z^2 \le 4-x^2
\]
For a fixed \(x\), the cross-section in the \(yz\)-plane is a square:
\[ - \sqrt{4-x^2} \le y \le \sqrt{4-x^2}, \quad - \sqrt{4-x^2} \le z \le \sqrt{4-x^2} \]
Side length of the square: \[ 2\sqrt{4-x^2} \]
Area of the cross-section: \[ A(x)=\left(2\sqrt{4-x^2}\right)^2=4(4-x^2) \]
Step 3: Set up the volume integral.
The bounds for \(x\) are determined by:
\[
x^2 \le 4 \;\Rightarrow\; -2 \le x \le 2
\]
Hence, \[ V=\int_{-2}^{2} 4(4-x^2)\,dx \]
Step 4: Evaluate the integral.
\[ V=4\int_{-2}^{2}(4-x^2)\,dx \]
\[ =4\left[4x-\frac{x^3}{3}\right]_{-2}^{2} \]
\[ =4\left[\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\right] \]
\[ =4\left(\frac{32}{3}\right)=\frac{128}{3} \]
Final Answer:
\[ \boxed{\dfrac{128}{3}\;\approx\;42.67} \]