Question

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

Answer 1

Answer: V = $\frac{34}{45} \pi$  

Explanation:

In the given system of coordinates OXY, the region R₃ is bounded by two functions:

y₁ = $2\sqrt[4]{x}$  (green line)

y₂ = 2x  (blu line)

in the intervals:

0 ≤ x ≤ 1

0 ≤ y ≤ 2

We need to find the volume of this region rotated about the line AB, which is x = 1. In order to do so, we need to change system of coordinates, such as the rotation is about the y-axis, therefore we need to perform a translation:

$\left \{ {{X=x+1} \atop {Y=y}} \right.$

After the translation R₃ will be bounded by:

y₁ = $2\sqrt[4]{x+1}$

y₂ = 2x + 2

in the intervals:

-1 ≤ x ≤ 0

0 ≤ y ≤ 2

At this point, we can use the washer method (see picture attached). The general formula is:

A = π(R² - r²)

where:

A = area

R = outer radius of a washer

r = inner radius of a washer

Since the radii are x-values which vary with the height, represented by the y-values, we need to write the inverse functions:

$R: x_{1} = \frac{1}{16} y^{4} - 1 \\ r: x_{2} = \frac{1}{2} y - 1$

[Note: I used the curves on the left side of the graph, but you could find the ones representing the right side of the graph and use those]

Now, we can find the function for the area of each washer:

$A(y) = \pi [(\frac{1}{16}y^{4} - 1)^{2} - (\frac{1}{2}y - 1)^{2} ] \\ = \pi [\frac{1}{256}y^{8} - \frac{1}{8} y^{4} - \frac{1}{4} y^{2} + y ]$

Therefore the volume of the region R₃ will be:

$V = \int\limits^{y_{2}}_{y_{1}} {A(y)} \, dy$

$= \int\limits^2_0 {\pi [\frac{1}{256}y^{8} - \frac{1}{8}y^{4} - \frac{1}{4} y^{2} + y] } \, dy $

$= \pi [ \frac{1}{2304}y^{9} - \frac{1}{40}y^{5} - \frac{1}{12} y^{3} + \frac{1}{2} y^{2}]^{2}_{0}$

$= \frac{34}{45} \pi$

Answer 2

The volume generated by the region $R_{3}$ about AB is $\boxed{\bf \dfrac{34\pi}{45}}$.

Further explanation:

Formula used:

The volume genreated by region $R_{3}$ about AB can be obtained by the formula:

$\boxed{V=\int\limits_a^b{\pi\left({{r_1}^2-r_2^2}\right)dy}}$        …… (1)

Here, $V$ is the volume of the region, $r_{1}$ is the outer radius and $r_{2}$ is the inner radius.

Calculation:

According to the Washer method integrate along the axis parallel to the Axis of the rotation.

Here, $R_{3}$ is rotated about AB, then the axis of the rotation is $x=1$.

The outer radius $r_{1}$ is the distance from the curve $x=\frac{y^{4}}{16}$ to the axis of rotaion $x=1$.

$\boxed{r_{1}= 1-\frac{{{y^4}}}{{16}}}$

The inner radius $r_{2}$ is the distance from the curve $x=y$ to tha axis of rotaion is $x=1$.

$\boxed{r_{2}=1-y }$

Substitute $(1-y)$ for $r_{2}$ and $1-\frac{y^{4}}{16}$ for $r_{1}$, $2$ for $b$ and $0$ for $a$ in equation (1) to obtain the volume generated by $R_{3}$ about AB.

$\begin{aligned}V&=\int\limits_0^2{\pi\left({{{\left({1-\frac{{{y^4}}}{{16}}}\right)}^2}-{{\left({1-y}\right)}^2}} \right)dy}\\&=\int\limits_0^2{\pi\left({\left({1+\frac{{{y^8}}}{{256}}-2\left({\frac{{{y^4}}}{{16}}}\right)1}\right)-\left({1+{y^2}-2y}\right)}\right)}dy\\&=\int\limits_0^2{\pi\left({\left( {1+\frac{{{y^8}}}{{256}}-\left({\frac{{{y^4}}}{8}}\right)}\right)-\left({1+{y^2}-2y}\right)} \right)}dy\end{aligned}$

Further solve the above equation as follows:

$\begin{aligned}V&=\int\limits_0^2{\pi\left({\left({\frac{{{y^8}}}{{256}}-\left({\frac{{{y^4}}}{8}}\right)}\right)- \left({{y^2}-2y}\right)}\right)}dy\\&=\pi\left[{\frac{{{y^9}}}{{256\left(9\right)}}-\frac{{{y^5}}}{{5\left(8\right)}}-\frac{{{y^3}}}{3}+\frac{{2{y^2}}}{2}}\right]_0^2\\&=\pi\left[{\frac{{{2^9}}}{{256\left(9 \right)}}-\frac{{{2^5}}}{{5\left(8\right)}}-\frac{{{2^3}}}{3}+4}\right]\\&=\pi\left[{\frac{2}{9}-\frac{4}{5}-\frac{8}{3}+4}\right]\\&=\dfrac{34\pi}{45}\end{aligned}$

Therefore, the volume generated by the region $R_{3}$ about AB is $\boxed{\bf \dfrac{34\pi}{45}}$.

Learn more:

1. Simplification: https://brainly.com/question/1602237

2. Quadratic equation: https://brainly.com/question/1332667

Answer details:

Grade: College

Subject: Mathematics

Chapter: Calculus

Keywords: Integration, volume, dy, inner radius, outer radius, rotation, axis, x-axis, y-axis, coordinate, generated by the curve.