Respuesta :

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.

Ver imagen virtuematane

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]

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function https://brainly.com/question/3412497

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.