Question

Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 4 into two regions with equal area. (Round your answer to two decimal places.) b =

Answer 1

Answer:

$b=\sqrt[3]{16}$

Step-by-step explanation:

Determine interception point of curves

$36x^2=4$

$x^2=\frac{4}{36}$

$x^2=\frac{1}{9}$

$x=\pm\frac{1}{3}$

Find the area of the bounded region

$\int\limits^\frac{1}{3} _{-\frac{1}{3}} {4-36x^2} \, dx\\ \\=4x-12x^3\Bigr|_{-\frac{1}{3} }^{\frac{1}{3} }\\\\=[4(\frac{1}{3})-12(\frac{1}{3})^3]-[4(-\frac{1}{3})-12(-\frac{1}{3})^3]\\\\=\frac{8}{9}-(-\frac{8}{9})\\\\= \frac{8}{9}+\frac{8}{9} \\\\=\frac{16}{9}$

Therefore, since half of the area is $\frac{8}{9}$, we can set one-half of the region between 0 and x where $x=\frac{\sqrt{b}}{6}$ and determine b:

$\frac{8}{9}=\int\limits^\frac{\sqrt{b}}{6} _0 {b-36x^2} \, dx\\\\\frac{8}{9} =bx-12x^3\Bigr|_{0}^{\frac{\sqrt{b}}{6} }\\\\\frac{8}{9}=b(\frac{\sqrt{b}}{6})-12(\frac{\sqrt{b}}{6})^3\\\\\frac{8}{9}=\frac{b\sqrt{b}}{6}-12(\frac{b^{\frac{3}{2}}}{216})\\\\\frac{8}{9}=\frac{b\sqrt{b}}{6}-(\frac{b^{\frac{3}{2}}}{18})\\\\\frac{16}{18}=\frac{3b\sqrt{b}}{18}-\frac{b^{\frac{3}{2}}}{18}\\\\16=3b\sqrt{b}-b^{\frac{3}{2}}\\\\16=3b^{\frac{3}{2}}-b^{\frac{3}{2}}\\\\16=2b^{\frac{3}{2}}$

To account for both halves of the region:

$16=2(2b^{\frac{3}{2}})\\16=4b^{\frac{3}{2}}\\4=b^{\frac{3}{2}}\\b=\sqrt[3]{16}$

Therefore, the line $b=\sqrt[3]{16}$ will divide the area between the curves into two regions with equal area