let $r$ be the region enclosed by $f(x)=x^{2}+4$, $g(x)=(x - 3)^{2}$ and the axes $x = 0$ and $y = 0$…

let $r$ be the region enclosed by $f(x)=x^{2}+4$, $g(x)=(x - 3)^{2}$ and the axes $x = 0$ and $y = 0$. compute the volume $v$ obtained by rotating $r$ about the $x$-axis. hint: express $v$ as a sum of two integrals. $v=$

let $r$ be the region enclosed by $f(x)=x^{2}+4$, $g(x)=(x - 3)^{2}$ and the axes $x = 0$ and $y = 0$. compute the volume $v$ obtained by rotating $r$ about the $x$-axis. hint: express $v$ as a sum of two integrals. $v=$

Answer

Explanation:

Step1: Encontrar el punto de intersección

Resolvemos $x^{2}+4=(x - 3)^{2}$. Expandimos $(x - 3)^{2}=x^{2}-6x + 9$. Entonces $x^{2}+4=x^{2}-6x + 9$, lo que da $6x=5$ y $x=\frac{5}{6}$.

Step2: Usar el método de discos

El volumen $V$ de la región rotada alrededor del eje $x$ está dado por $V=\pi\int_{a}^{b}[f(x)^{2}]dx$. La región $R$ se divide en dos partes. Para $0\leq x\leq\frac{5}{6}$, la función es $y = x^{2}+4$, y para $\frac{5}{6}\leq x\leq3$, la función es $y=(x - 3)^{2}$. El volumen $V$ es la suma de dos integrales: $V=\pi\int_{0}^{\frac{5}{6}}(x^{2}+4)^{2}dx+\pi\int_{\frac{5}{6}}^{3}(x - 3)^{4}dx$

Step3: Expandir y integrar

Expansión de $(x^{2}+4)^{2}=x^{4}+8x^{2}+16$. $\int(x^{4}+8x^{2}+16)dx=\frac{1}{5}x^{5}+\frac{8}{3}x^{3}+16x+C$. Evaluando $\pi\int_{0}^{\frac{5}{6}}(x^{2}+4)^{2}dx=\pi\left[\frac{1}{5}\left(\frac{5}{6}\right)^{5}+\frac{8}{3}\left(\frac{5}{6}\right)^{3}+16\left(\frac{5}{6}\right)\right]$. Expansión de $(x - 3)^{4}=x^{4}-12x^{3}+54x^{2}-108x + 81$. $\int(x - 3)^{4}dx=\frac{1}{5}(x - 3)^{5}+C$. Evaluando $\pi\int_{\frac{5}{6}}^{3}(x - 3)^{4}dx=\pi\left[0-\frac{1}{5}\left(\frac{5}{6}-3\right)^{5}\right]$. Sumando ambas integrales: [ \begin{align*} V&=\pi\left[\frac{1}{5}\left(\frac{5}{6}\right)^{5}+\frac{8}{3}\left(\frac{5}{6}\right)^{3}+16\left(\frac{5}{6}\right)+\frac{1}{5}\left(\frac{13}{6}\right)^{5}\right]\ &=\pi\left[\frac{5^{5}}{5\times6^{5}}+\frac{8\times5^{3}}{3\times6^{3}}+\frac{80}{6}+\frac{13^{5}}{5\times6^{5}}\right]\ &=\pi\left[\frac{3125}{5\times7776}+\frac{8\times125}{3\times216}+\frac{80}{6}+\frac{371293}{5\times7776}\right]\ &=\pi\left[\frac{3125 + 371293}{5\times7776}+\frac{1000}{648}+\frac{80}{6}\right]\ &=\pi\left[\frac{374418}{38880}+\frac{1000}{648}+\frac{80}{6}\right]\ &=\pi\left[\frac{62403}{6480}+\frac{1000}{648}+\frac{8640}{648}\right]\ &=\pi\left[\frac{62403+1000 + 8640}{6480}\right]\ &=\frac{72043\pi}{6480} \end{align*} ]

Answer:

$\frac{72043\pi}{6480}$