Let \( S \) be the solid obtained by rotating the region bounded by the curves \( y=\sin \left(x^{2}\right) \) and \( y=0 \), with \( 0 \leq x \leq \sqrt{\pi} \), about the \( y \) raxis. Use cylindrical shells to find the volume of \( S \). Volume \( = \)

To find the volume of the solid \( S \) obtained by rotating the region bounded by the curves \( y = \sin(x^2) \) and \( y = 0 \) for \( 0 \leq x \leq \sqrt{\pi} \) about the \( y \)-axis using the method of cylindrical shells, we can use the formula for the volume of a solid of revolution given by cylindrical shells: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( f(x) \) is the height of the shell, and \( a \) and \( b \) are the bounds of integration. In our case, the height of the shell is given by \( f(x) = \sin(x^2) \), and the bounds are \( a = 0 \) and \( b = \sqrt{\pi} \). Thus, we can set up the integral as follows: \[ V = 2\pi \int_{0}^{\sqrt{\pi}} x \cdot \sin(x^2) \, dx \] To evaluate this integral, we can use the substitution method. Let: \[ u = x^2 \quad \Rightarrow \quad du = 2x \, dx \quad \Rightarrow \quad dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}} \] When \( x = 0 \), \( u = 0 \), and when \( x = \sqrt{\pi} \), \( u = \pi \). Therefore, the limits of integration change from \( 0 \) to \( \pi \). Substituting into the integral, we have: \[ V = 2\pi \int_{0}^{\pi} x \cdot \sin(x^2) \cdot \frac{du}{2\sqrt{u}} = \pi \int_{0}^{\pi} \sin(u) \, du \] Now we can evaluate the integral \( \int_{0}^{\pi} \sin(u) \, du \): \[ \int \sin(u) \, du = -\cos(u) + C \] Thus, \[ \int_{0}^{\pi} \sin(u) \, du = \left[-\cos(u)\right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \] Now substituting back into the volume formula: \[ V = \pi \cdot 2 = 2\pi \] Therefore, the volume of the solid \( S \) is: \[ \boxed{2\pi} \]