Question

the answer above is not correct. (1 point) find the volume of the solid formed by rotating the region enclosed by x = 0, x = 1, y = 0, y = 9 + x^9 about the x - axis. volume = 436π/5 you have attempted this problem 1 time. you received a score of 0% for this attempt. your overall recorded score is 0%. your score was recorded. your score was successfully sent to the lms preview my answers submit answers

Explanation:

Step1: Recall volume - of - revolution formula

The volume $V$ of the solid formed by rotating the region bounded by $y = f(x)$, $x=a$, $x = b$ and $y = 0$ about the $x$ - axis is given by $V=\pi\int_{a}^{b}[f(x)]^{2}dx$. Here, $a = 0$, $b = 1$, and $f(x)=9 + x^{3}$.

Step2: Square the function

$[f(x)]^{2}=(9 + x^{3})^{2}=81+18x^{3}+x^{6}$ using the formula $(A + B)^{2}=A^{2}+2AB + B^{2}$ where $A = 9$ and $B=x^{3}$.

Step3: Integrate term - by - term

$\int_{0}^{1}(81+18x^{3}+x^{6})dx=\int_{0}^{1}81dx+\int_{0}^{1}18x^{3}dx+\int_{0}^{1}x^{6}dx$.
$\int_{0}^{1}81dx=81x\big|_{0}^{1}=81(1 - 0)=81$.
$\int_{0}^{1}18x^{3}dx=18\times\frac{x^{4}}{4}\big|_{0}^{1}=\frac{18}{4}(1 - 0)=\frac{9}{2}$.
$\int_{0}^{1}x^{6}dx=\frac{x^{7}}{7}\big|_{0}^{1}=\frac{1}{7}(1 - 0)=\frac{1}{7}$.
Then $\int_{0}^{1}(81+18x^{3}+x^{6})dx=81+\frac{9}{2}+\frac{1}{7}$.
Find a common denominator, which is $14$.
$81=\frac{81\times14}{14}=\frac{1134}{14}$, $\frac{9}{2}=\frac{9\times7}{14}=\frac{63}{14}$, $\frac{1}{7}=\frac{2}{14}$.
So $\int_{0}^{1}(81+18x^{3}+x^{6})dx=\frac{1134 + 63+2}{14}=\frac{1199}{14}$.

Step4: Multiply by $\pi$

$V=\pi\int_{0}^{1}(9 + x^{3})^{2}dx=\frac{1199\pi}{14}$

Answer:

$\frac{1199\pi}{14}$