find the volume of the solid obtained by rotating the region in the first quadrant bounded by ( y = x^{6}, y…

Answer:

$\frac{12\pi}{13}$

Explanation:

Step1: Find the intersection point

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

Step2: Use the disk method formula

The formula for the volume of a solid of revolution using the disk method when rotating about the $x$-axis is $V=\pi\int_{a}^{b}[R(x)]^{2}dx$. Here, $R(x)=1$ (outer radius) and $r(x)=x^{6}$ (inner radius, since we are rotating the region between $y = 1$ and $y=x^{6}$). The limits of integration are from $x = 0$ to $x = 1$. So $V=\pi\int_{0}^{1}(1^{2}-(x^{6})^{2})dx=\pi\int_{0}^{1}(1 - x^{12})dx$.

Step3: Integrate

Integrate term - by - term. $\int(1 - x^{12})dx=x-\frac{x^{13}}{13}+C$.

Step4: Evaluate the definite integral

$V=\pi\left[x-\frac{x^{13}}{13}\right]_{0}^{1}$. Substitute the upper and lower limits: $V=\pi\left(1-\frac{1}{13}-(0 - 0)\right)=\pi\left(\frac{13 - 1}{13}\right)=\frac{12\pi}{13}$.