Chapter 6.1, Problem 41E

To determine

To find the value of $b$ when the line $y=b$ divides the region bounded by the curves into two regions. With equal area.

Answer to Problem 41E

The value of $b=\sqrt[3]{16}$

Explanation of Solution

Given: The line $y=b$

Calculation:

  

  $\begin{array}{l}y=x^2\text{ ---(1)}\\ \text{y = 4 ---(2)}\\ \text{From 1 and 2 }\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm 2\\ If\text{ x=2, y=4 and x=-2, y=4}\\ \text{The two curves cut at (2,4) \& (-2,4)}\\ y=x^2\text{ is a parabola as 'y' as axis}\text{.}\\ \Rightarrow \text{Area between 1 and 2 is equal to A}\text{.}\\ \Rightarrow A=2{\displaystyle \underset{0}{\overset{4}{\int }}x\text{ dy}}\\ =2{\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{y}\text{ dy}}\\ =2{\left(\frac{y^{3/2}}{3/2}\right)}_0^4\\ =2\ast 2/3(4^{3/2})\\ 32/3\\ 10.66\\ Area\text{ between 1 and y = b is }\frac{1}{2}\text{(32/3)}\\ \text{=16/3}\text{.(Because y=b divides area into equal parts)}\\ \Rightarrow 2{\displaystyle \underset{0}{\overset{b}{\int }}\sqrt{y}\text{ dy}}=16/3\\ \Rightarrow {\left(\frac{y^{3/2}}{3/2}\right)}_0^b=16/3\\ \Rightarrow \frac{2}{3}{b}^{3/2}=16/3\\ \Rightarrow b^{3/2}=4\\ \Rightarrow b=\sqrt[3]{16}\\ Hence,the\text{ value of b = }\sqrt[3]{16}.\end{array}$

Hence, the value of b $=\sqrt[3]{16}sq.units$