Suppose \( f(x)>0 \) for all \( x \) and \( \int_{0}^{49} f(x) d x=16 \). Let \( R \) be the region in the first quadrant bounded by the coordinates axes, \( y=f\left(x^{2}\right) \), and \( x=7 \). Find the volume of the solid generated by revolving \( R \) around the \( y \)-axis.

To find the volume of the solid generated by revolving the region \( R \) around the \( y \)-axis, we can use the method of cylindrical shells. The region \( R \) is bounded by the \( y \)-axis, the line \( x=7 \), and the curve \( y=f(x^2) \). The formula for the volume \( V \) of the solid of revolution using cylindrical shells is given by: \[ V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx \] where \( h(x) \) is the height of the shell at position \( x \). In our case, the limits of integration are from \( x=0 \) to \( x=7 \), and the height of each shell is \( h(x) = f(x^2) \). Therefore, we can write the volume \( V \) as follows: \[ V = 2\pi \int_{0}^{7} x f(x^2) \, dx \] Next, we perform the substitution \( u = x^2 \), which gives \( du = 2x \, dx \) or \( dx = \frac{du}{2\sqrt{u}} \). When \( x=0 \), \( u=0 \), and when \( x=7 \), \( u=49 \). Therefore, we can change the limits of integration accordingly: \[ V = 2\pi \int_{0}^{49} x f(x^2) \, dx = 2\pi \int_{0}^{49} \sqrt{u} f(u) \cdot \frac{du}{2\sqrt{u}} \] The \( \sqrt{u} \) cancels out: \[ V = \pi \int_{0}^{49} f(u) \, du \] We know from the problem statement that: \[ \int_{0}^{49} f(x) \, dx = 16 \] Thus, substituting this result in the expression for volume, we have: \[ V = \pi \cdot 16 = 16\pi \] Therefore, the volume of the solid generated by revolving the region \( R \) around the \( y \)-axis is \[ \boxed{16\pi}. \]