volumes by cross - section lecture participation: problem 1\n(2 points)\nfind the volume of the solid whose…

Explanation:

Step1: Find the limits of integration

Set (y = x^{6}) equal to (y = 1). Then (x^{6}=1), and since (x\geq0) (first - quadrant), (x = 1). The limits of integration are from (x = 0) to (x = 1).

Step2: Find the diameter of the semic - circle

The base of each semic - circular cross - section perpendicular to the (x) - axis is (b=1 - x^{6}). The radius of the semic - circle (r=\frac{1 - x^{6}}{2}).

Step3: Find the area of the semic - circle

The area of a semic - circle is (A=\frac{1}{2}\pi r^{2}). Substituting (r=\frac{1 - x^{6}}{2}), we get (A(x)=\frac{\pi}{2}\left(\frac{1 - x^{6}}{2}\right)^{2}=\frac{\pi}{8}(1 - 2x^{6}+x^{12})).

Step4: Integrate the area function

The volume (V=\int_{a}^{b}A(x)dx). Here, (a = 0), (b = 1), and (A(x)=\frac{\pi}{8}(1 - 2x^{6}+x^{12})). [ \begin{align*} V&=\frac{\pi}{8}\int_{0}^{1}(1 - 2x^{6}+x^{12})dx\ &=\frac{\pi}{8}\left[\int_{0}^{1}1dx-2\int_{0}^{1}x^{6}dx+\int_{0}^{1}x^{12}dx\right]\ \end{align*} ] Using the power rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have: [ \begin{align*} \int_{0}^{1}1dx&=x\big|{0}^{1}=1-0 = 1\ 2\int{0}^{1}x^{6}dx&=2\times\frac{x^{7}}{7}\big|{0}^{1}=\frac{2}{7}\ \int{0}^{1}x^{12}dx&=\frac{x^{13}}{13}\big|_{0}^{1}=\frac{1}{13} \end{align*} ] [ \begin{align*} V&=\frac{\pi}{8}\left(1-\frac{2}{7}+\frac{1}{13}\right)\ &=\frac{\pi}{8}\left(\frac{91-26 + 7}{91}\right)\ &=\frac{\pi}{8}\times\frac{72}{91}\ &=\frac{9\pi}{91} \end{align*} ]

Answer:

(\frac{9\pi}{91})