Question

1. let ( r ) be the triangular region in the first quadrant, with vertices at points ( (0,0) ), ( (0,2) ), and ( (1,2) ). the region ( r ) is the base of a solid. for the solid, each cross section perpendicular to the ( y )-axis is an isosceles right triangle with the right angle on the ( y )-axis and one leg in the ( xy )-plane. what is the volume of the solid? (a) ( \frac{1}{3} ) (b) ( \frac{1}{2} ) (c) ( \frac{2}{3} ) (d) ( \frac{16}{3} )

Explanation:

Step1: Find the equation of the line

The line passing through \((0,0)\) and \((1,2)\) has the equation \(x=\frac{y}{2}\) (using the slope - intercept form \(y = mx + b\), where \(m = 2\) and \(b = 0\), then solving for \(x\)).

Step2: Determine the area of the cross - section

Since each cross - section perpendicular to the \(y\) - axis is an isosceles right triangle with one leg \(l=x\) (in the \(xy\) - plane). The area of an isosceles right triangle \(A=\frac{1}{2}l^{2}\). Substituting \(l = x=\frac{y}{2}\), we get \(A(y)=\frac{1}{2}(\frac{y}{2})^{2}=\frac{y^{2}}{8}\).

Step3: Set up the volume integral

The limits of integration for \(y\) are from \(y = 0\) to \(y = 2\). Using the formula for the volume of a solid with known cross - sectional area \(V=\int_{a}^{b}A(y)dy\), we have \(V=\int_{0}^{2}\frac{y^{2}}{8}dy\).

Step4: Evaluate the integral

Using the power rule \(\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C(n
eq - 1)\), we get \(V=\frac{1}{8}\times\frac{y^{3}}{3}\big|_{0}^{2}\).

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Answer:

A. \(\frac{1}{3}\)