find the area of the region enclosed by the curves. y = |x^2 - 36| and y = x^2/9 + 36. the area is…

Answer:

$2304$

Explanation:

Step1: Analyze $y = |x^{2}-36|$

$y=\begin{cases}x^{2}-36, &x\leq - 6\text{ or }x\geq6\36 - x^{2},&-6<x<6\end{cases}$

Step2: Find intersection points

Set $|x^{2}-36|=\frac{x^{2}}{9}+36$. Case 1: When $x\leq - 6$ or $x\geq6$, $x^{2}-36=\frac{x^{2}}{9}+36$. $x^{2}-\frac{x^{2}}{9}=72$, $\frac{8x^{2}}{9}=72$, $x^{2}=81$, $x=\pm9$. Case 2: When $-6 < x < 6$, $36 - x^{2}=\frac{x^{2}}{9}+36$. $-\ x^{2}-\frac{x^{2}}{9}=0$, $-\frac{10x^{2}}{9}=0$, $x = 0$.

Step3: Set up integral for area

$A=\int_{-9}^{-6}\left[\left(\frac{x^{2}}{9}+36-(x^{2}-36)\right)\right]dx+\int_{-6}^{6}\left[\left(\frac{x^{2}}{9}+36-(36 - x^{2})\right)\right]dx+\int_{6}^{9}\left[\left(\frac{x^{2}}{9}+36-(x^{2}-36)\right)\right]dx$. Since the functions are even, $A = 2\int_{0}^{6}\left(\frac{10x^{2}}{9}\right)dx+2\int_{6}^{9}\left(72-\frac{8x^{2}}{9}\right)dx$.

Step4: Integrate first part

$\int_{0}^{6}\frac{10x^{2}}{9}dx=\frac{10}{9}\times\frac{x^{3}}{3}\big|_{0}^{6}=\frac{10}{27}\times216 = 80$.

Step5: Integrate second part

$\int_{6}^{9}\left(72-\frac{8x^{2}}{9}\right)dx=\left(72x-\frac{8x^{3}}{27}\right)\big|_{6}^{9}=(72\times9-\frac{8\times9^{3}}{27})-(72\times6-\frac{8\times6^{3}}{27})=(648 - 216)-(432-\frac{8\times216}{27})=(432)-(432 - 64)=64$.

Step6: Calculate total area

$A = 2\times80+2\times64\times3=160 + 384\times2=160+768 = 2304$.