Mathematics

Problem

A student is asked to find the volume of a solid whose cross sections perpendicular to the

y

-axis are squares. The solid sits on the region in the first quadrant bounded by the

y

-axis, the

x

-axis, and the curve

y=6-x^{2}

. The student proposes the following integral:

V=\int_{0}^{6} \left(6-y\right)^{2} \, dy.

(a) Identify and explain every error in the student’s integral set-up. (b) Write the corrected definite integral that represents the volume of the solid, justifying each component of your expression (integrand, limits, and differential).

Intuitive explanation

The student has confused which variable describes the side length of the square cross-sections. Since cross-sections are perpendicular to the $y$ -axis, you need to express the side of each square as a function of $y$ —that is, solve $y = 6 - x^2$ for $x$ in terms of $y$ . The expression $6 - y$ gives the $x$ -coordinate on the curve, but that's only half the story: the square's side length is the full width across the region at height $y$ , which is $2x = 2\sqrt{6-y}$ . Additionally, the upper limit should reflect where the parabola actually reaches, not an arbitrary endpoint.

Intuitive explanation

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