Refer to the figure and find the volume V generated by rotating the given region about the specified line.
R3 about AB

Answer:
Hence, volume is: [tex]\dfrac{34\pi}{45}[/tex] cubic units.
Step-by-step explanation:
We will first express our our equation of the curve and the line bounded by the region in terms of the variable y.
i.e. the curve is re[tex]x=\dfrac{1}{16}y^4[/tex]
and the line is given as: [tex]x=\dfrac{1}{2}y[/tex]
Since after rotating the given region [tex]R_{3}[/tex] about the line AB.
we see that for the following graph
the axis is located at x=1.
and the outer radius(R) is: [tex]\dfrac{1}{16}y^4[/tex]
and the inner radius(r) is: [tex]\dfrac{1}{2}y[/tex]
Now, the area of the graph= area of the disc.
Area of graph=[tex]\pi(R^2-r^2)[/tex]
Now the volume is given as:
[tex]Volume=\int\limits^2_0 {Area} \, dy[/tex]
On calculating we get:
Volume=[tex]\dfrac{34\pi}{45}[/tex] cubic units.
The volume V generated by rotating the given region about the specified line R3 about AB is [tex]\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.[/tex]
Further explanation:
Given:
The coordinates of point A is [tex]\left( {1,0} \right).[/tex]
The coordinates of point B is [tex]\left( {1,2} \right).[/tex]
The coordinate of point C is [tex]\left( {0,2} \right).[/tex]
The value of y is [tex]y = 2\sqrt[4]{x}.[/tex]
Explanation:
The equation of the curve is [tex]y = 2\sqrt[4]{x}.[/tex]
Solve the above equation to obtain the value of x in terms of y.
[tex]\begin{aligned}{\left( y \right)^4}&={\left( {2\sqrt[4]{x}} \right)^4} \\{y^4}&=16x\\\frac{1}{{16}}{y^4}&= x\\\end{aligned}[/tex]
The equation of the line is [tex]x = \dfrac{1}{2}y.[/tex]
After rotating the region [tex]{R_3}[/tex] is about the line AB.
From the graph the inner radius is [tex]{{r_2} = \dfrac{1}{2}y[/tex] and the outer radius is [tex]{{r_1}=\dfrac{1}{{16}}{y^4}.[/tex]
[tex]{\text{Area of graph}}=\pi\left( {{r_1}^2 - {r_2}^2} \right)[/tex]
[tex]Area = \pi\left( {{{\left({\frac{1}{{16}}{y^4}} \right)}^2} - {{\left({\frac{1}{2}y} \right)}^2}}\right)[/tex]
The volume can be obtained as follows,
[tex]\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}[/tex]
Further solve the above equation.
[tex]\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}[/tex]
The volume V generated by rotating the given region about the specified line R3 about AB is [tex]\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.[/tex]
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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.