volumes by disks: problem 2\n(1 point)\nfind the volume of the solid obtained by rotating the region in the…

Answer:

$\frac{1016\pi}{63}$

Explanation:

Step1: Find intersection point

Set $y = x^{6}=1$, then $x = 1$ (since in the first - quadrant).

Step2: Use the washer method formula

The formula for the volume $V=\pi\int_{a}^{b}[(R(x))^{2}-(r(x))^{2}]dx$ when rotating about a horizontal line. Here, $a = 0$, $b = 1$, $R(x)=1 + 5=6$ (distance from $y=-5$ to $y = 1$) and $r(x)=x^{6}+5$ (distance from $y=-5$ to $y=x^{6}$). So $V=\pi\int_{0}^{1}[6^{2}-(x^{6}+5)^{2}]dx$.

Step3: Expand the integrand

Expand $(x^{6}+5)^{2}=x^{12}+10x^{6}+25$. Then $[6^{2}-(x^{6}+5)^{2}]=36-(x^{12}+10x^{6}+25)=11 - x^{12}-10x^{6}$.

Step4: Integrate term - by - term

$\int_{0}^{1}(11 - x^{12}-10x^{6})dx=\int_{0}^{1}11dx-\int_{0}^{1}x^{12}dx-\int_{0}^{1}10x^{6}dx$. Using $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: $\int_{0}^{1}11dx=11x|{0}^{1}=11$, $\int{0}^{1}x^{12}dx=\frac{x^{13}}{13}|{0}^{1}=\frac{1}{13}$, $\int{0}^{1}10x^{6}dx=10\times\frac{x^{7}}{7}|_{0}^{1}=\frac{10}{7}$.

Step5: Calculate the volume

$V=\pi\left(11-\frac{1}{13}-\frac{10}{7}\right)=\pi\left(\frac{11\times91 - 7-130}{91}\right)=\pi\left(\frac{1001-7 - 130}{91}\right)=\pi\left(\frac{1016}{63}\right)$.