The volume V generated by rotating the given region about the specified line R3 about AB is \(\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.\)
Further explanation:
Given:
The coordinates of point A is \(\left( {1,0} \right).\)
The coordinates of point B is \(\left( {1,2} \right).\)
The coordinate of point C is \(\left( {0,2} \right).\)
The value of y is \(y = 2\sqrt[4]{x}.\)
Explanation:
The equation of the curve is \(y = 2\sqrt[4]{x}.\)
Solve the above equation to obtain the value of x in terms of y.
\(\begin{aligned}{\left( y \right)^4}&={\left( {2\sqrt[4]{x}} \right)^4} \\{y^4}&=16x\\\frac{1}{{16}}{y^4}&= x\\\end{aligned}\)
The equation of the line is \(x = \dfrac{1}{2}y.\)
After rotating the region \({R_3}\) is about the line AB.
From the graph the inner radius is \({{r_2} = \dfrac{1}{2}y\) and the outer radius is \({{r_1}=\dfrac{1}{{16}}{y^4}.\)
\({\text{Area of graph}}=\pi\left( {{r_1}^2 - {r_2}^2} \right)\)
\(Area = \pi\left( {{{\left({\frac{1}{{16}}{y^4}} \right)}^2} - {{\left({\frac{1}{2}y} \right)}^2}}\right)\)
The volume can be obtained as follows,
\(\begin{aligned}{\text{Volume}}&=\int\limits_0^2 {Area{\text{ }}dy}\\&=\int\limits_0^2{\pi \left( {{{\left({\frac{1}{{16}}{y^4}} \right)}^2} - {{\left( {\frac{1}{2}y} \right)}^2}} \right){\text{ }}dy}\\&= \pi \int\limits_0^2 {\left( {\frac{1}{{256}}{y^8} - \frac{1}{4}{y^2}} \right){\text{ }}dy}\\\end{aligned}\)
Further solve the above equation.
\(\begin{aligned}{\text{Volume}}&=\pi \left[ {\int\limits_0^2 {\frac{1}{{256}}{y^8}dy - } \int\limits_0^2{\frac{1}{4}{y^2}{\text{ }}dy} } \right]\\&= \frac{{34\pi }}{{45}}\\\end{aligned}\)
The volume V generated by rotating the given region about the specified line R3 about AB is \(\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.\)
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Volume of the curves
Keywords: area, volume of the region, rotating, generated, specified line, R3, AB, rotating region.