Question

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

Answer 1

Answer:

Hence, volume is: $\dfrac{34\pi}{45}$ 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$x=\dfrac{1}{16}y^4$

and the line is given as:  $x=\dfrac{1}{2}y$

Since after rotating the given region $R_{3}$ about the line AB.

we see that for the following graph

the axis is located at x=1.

and the outer radius(R) is: $\dfrac{1}{16}y^4$

and the inner radius(r) is:  $\dfrac{1}{2}y$

Now, the area of the graph= area of the disc.

Area of graph=$\pi(R^2-r^2)$

Now the volume is given as:

$Volume=\int\limits^2_0 {Area} \, dy$

On calculating we get:

Volume=$\dfrac{34\pi}{45}$ cubic units.

Answer 2

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 $\boxed{\frac{{34\pi }}{{45}}{\text{ uni}}{{\text{t}}^3}}.$

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 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.