A solid is formed with triangular cross-sections where each triangle's base is defined by the equation \(y = x^2\) on interval \([0, 2]\). What is the volume of this solid?

To find the volume of the solid formed with triangular cross-sections defined by the equation \(y = x^2\) on the interval \([0, 2]\), we can use integration. First, we need to determine the area of each triangular cross-section. For a triangle, the area \(A\) is given by: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] In this case, the base of the triangle is defined by the line segment from \(y = 0\) to \(y = x^2\), which means the height of the triangle is the vertical distance from the x-axis to the curve, that is \(height = x^2\). For a triangle situated at a point \(x\) with base along the line \(y=x^2\), the base \(b\) of the triangle is also equal to \(x^2\), as it spans from the curve to the x-axis. Thus, we find: - Base of the triangle \(b = x^2\) - Height of the triangle \(h = x^2\) Thus, the area \(A(x)\) of the triangle as a function of \(x\) is: \[ A(x) = \frac{1}{2} \times b \times h = \frac{1}{2} \times x^2 \times x^2 = \frac{1}{2} x^4 \] To find the volume \(V\) of the solid, we now integrate the area function from \(x = 0\) to \(x = 2\): \[ V = \int_0^2 A(x) \, dx = \int_0^2 \frac{1}{2} x^4 \, dx \] Now calculating the integral: \[ V = \frac{1}{2} \int_0^2 x^4 \, dx \] The integral of \(x^4\) is: \[ \int x^4 \, dx = \frac{x^5}{5} \] Applying the bounds \(0\) to \(2\): \[ V = \frac{1}{2} \left[ \frac{x^5}{5} \right]_0^2 = \frac{1}{2} \left[ \frac{2^5}{5} - 0 \right] = \frac{1}{2} \left[ \frac{32}{5} \right] = \frac{16}{5} \] Thus, the volume of the solid is: \[ \boxed{\frac{16}{5}} \]