A solid has a cross section perpendicular to the x-axis, which is a square with side length \(y\). If the function \(y = 2x + 1\) represents the upper boundary of the solid from \(x=0\) to \(x=3\), find its volume.

To find the volume of the solid with square cross-sections perpendicular to the x-axis, we start by determining the side length of the square \(y\), which is given by the function \(y = 2x + 1\) for the interval \(x = 0\) to \(x = 3\). The area \(A\) of each square cross-section is given by: \[ A = y^2 = (2x + 1)^2. \] We need to compute the volume \(V\) of the solid by integrating the area of the squares along the x-axis from \(0\) to \(3\): \[ V = \int_{0}^{3} A \, dx = \int_{0}^{3} (2x + 1)^2 \, dx. \] Next, we expand the integrand: \[ (2x + 1)^2 = 4x^2 + 4x + 1. \] Now we can write the integral: \[ V = \int_{0}^{3} (4x^2 + 4x + 1) \, dx. \] We can break this up into separate integrals: \[ V = \int_{0}^{3} 4x^2 \, dx + \int_{0}^{3} 4x \, dx + \int_{0}^{3} 1 \, dx. \] Calculating each integral separately: 1. For \(\int_{0}^{3} 4x^2 \, dx\): \[ \int 4x^2 \, dx = \frac{4}{3}x^3 \bigg|_{0}^{3} = \frac{4}{3}(3^3) - \frac{4}{3}(0^3) = \frac{4}{3}(27) = 36. \] 2. For \(\int_{0}^{3} 4x \, dx\): \[ \int 4x \, dx = 2x^2 \bigg|_{0}^{3} = 2(3^2) - 2(0^2) = 2(9) = 18. \] 3. For \(\int_{0}^{3} 1 \, dx\): \[ \int 1 \, dx = x \bigg|_{0}^{3} = 3 - 0 = 3. \] Now we combine these results: \[ V = 36 + 18 + 3 = 57. \] Thus, the volume of the solid is: \[ \boxed{57}. \]