Chapter 8.2, Problem 43E

To determine

To Find:

To find limit of the ratio of the area of the triangle to the area of the parabolic region as $a$ approach zero.

Answer to Problem 43E

The ratio of the triangle to the parabola is $\frac{3}{4}$

Explanation of Solution

Given information:

The triangle AOC inscribed in the region cut from the parabola $y=x^2$ by the line $y=a^2$

  

The red line is the graph of $y=a^2$ , the blue curve is the graph of $y=x^2$ .

  

Now to find the area of the triangle,

Area of the triangle $=\frac{bh}{2}$

  $\begin{array}{l}=\frac{2a\cdot a^2}{2}\\ =a^3\end{array}$

Now to find the area of the parabola,

Area of the parabola $={\displaystyle \underset{-a}{\overset{a}{\int }}\left(a^2-x^2\right)}dx$

  $\begin{array}{l}={\left[a^{2}x-\frac{x^3}{3}\right]}_{-a}^a\\ =\left(a^3-\frac{a^3}{3}\right)-\left(-a^3+\frac{a^3}{3}\right)\\ =2a^3-\frac{2a^3}{3}\\ =\frac{6a^3-2a^3}{3}\\ =\frac{4a^3}{3}\end{array}$

Therefore, the ratio of the area of the triangle to the area of the parabolic region is, $\frac{a^3}{\frac{4a^3}{3}}=a^3\cdot \frac{3}{4a^3}=\frac{3}{4}$

Therefore, the ratio of the area of the triangle to the area of the parabolic region is $\frac{3}{4}$ .