Verified Answer :

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.

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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.