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?

We wish to find the volume of a solid whose cross‐sections (taken perpendicular to a given direction) are triangles. In our case the solid is “built on” a region in the xy–plane whose “base” is defined by the parabola   y = x²   for x between 0 and 2. A common interpretation of a problem of this type is as follows. Imagine the following setup:  • The “base” (or “footprint”) of the solid is the region in the plane bounded by the parabola y = x² and the x–axis, for x between 0 and 2.  • For every x in [0,2] a cross–section perpendicular to the x–axis is taken. On the interval from the x–axis up to the parabola the distance is   base length = (y on the parabola) – (y on the x–axis) = x² – 0 = x².  • On each such “base segment” a triangle is constructed. (In many such problems the cross–sections are taken to be “similar” — that is, of the same shape for all x. A common choice is that the triangle is equilateral.) In our problem we interpret the phrase “triangular cross–sections” to mean that the cross–section at a given x is an equilateral triangle with one side (its base) being the segment with length x². (If a different triangle were intended, one would need more information about the triangle—for example, whether it is isosceles with a given height. Since no extra info is given, the usual choice is equilateral.) Step 1. Write the Area of an Equilateral Triangle in Terms of Its Side Length For an equilateral triangle with side length s the area is given by   A = (√3/4)·s². Since in our problem the side length is the base of the triangle and equals x², the area of the cross–section at x is   A(x) = (√3/4)·(x²)² = (√3/4)·x⁴. Step 2. Set Up the Integral for the Volume The volume V of the solid is found by “stacking” these cross–sectional areas along the x–axis from x = 0 to x = 2. In other words,   V = ∫₀² A(x) dx    = ∫₀² (√3/4)x⁴ dx. Step 3. Evaluate the Integral Factor the constant √3/4 outside the integral:   V = (√3/4) ∫₀² x⁴ dx. Recall that   ∫ x⁴ dx = x⁵/5, so that   ∫₀² x⁴ dx = [x⁵/5]₀² = (2⁵/5) – (0⁵/5) = (32/5). Thus,   V = (√3/4)·(32/5) = (32√3)/(20) = (8√3)/5. Step 4. State the Final Answer The volume of the solid is   V = (8√3)/5  cubic units. Note: If the triangles were meant to be of another type (say, isosceles with a different relation between base and height) then additional information would be needed. However, in many standard problems the triangles are equilateral when no other information is provided. So, the answer is:   Volume = 8√3/5.