Find the volume of the solid generated by revolving the region bounded by y = 4*sqrt(x) and the...

Question:

Find the volume of the solid generated by revolving the region bounded by {eq}\; y = 4 \sqrt{x} \; {/eq} and the lines {eq}\; y = 4 \sqrt{3} \; {/eq} and {eq}\; x = 0 \; {/eq} about

A) the {eq}x {/eq}-axis.

B) the {eq}y {/eq}-axis.

C) the line {eq}y = 4 \sqrt{3} {/eq}.

D) the line {eq}x = 3 {/eq}.

Volumes of Solids of Revolution:

Given a region {eq}R {/eq} in the xy-plane bounded by the continuous functions {eq}F(x) \ge G(x) {/eq} over the interval {eq}a \le x \le b {/eq}.

Using thin cylindrical shells and thin disks/washers, the volumes of the solids generated when the region {eq}R {/eq} is revolved about some external axes are given by the integrals:

\begin{align*} &Volume = 2\pi \int_a^b (x - h) [ F(x) - G(x) ] \,dx & \text{[Volume of Revolution about the Vertical Line } x = h \text{ using Shell Method]} \\ &Volume = \pi \int_a^b \left( [F(x) - k]^2 - [G(x) - k]^2 \right) \,dx & \text{[Volume of Revolution about the Horizontal Line } y = k \text{ using Disk/Washer Method]} \end{align*}

Note that both methods assume the region {eq}R {/eq} to lie on the positive side of the axis of revolution, i.e., the region is either on the right or above the axis.

Otherwise, the volume of the generated solid of revolution is equal to the negative of these integrals.

Let {eq}R {/eq} be the region bounded by {eq}y = 4 \sqrt{x} \,,\, y = 4 \sqrt{3} \text{ and } x = 0 {/eq} as shown below:

Based on the graph, the bounded region can be defined as {eq}R = \left\{\, (x,y) \,|\, 0 \le x \le 3 \text{ and } 4 \sqrt{x} \le y \le 4 \sqrt{3} \,\right\} {/eq}.

A. Volume of Revolution about the X-Axis using the Disk/Washer Method

{eq}\begin{align*} Volume &= \pi \int_0^3 \left[ \left( 4 \sqrt{3} \right)^2 - \left( 4 \sqrt{x} \right)^2 \right] \,dx & \text{[The } x \text{-axis is also the horizontal line } y = 0 ] \\ &= \pi \int_0^3 16 (3 - x) \,dx & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= 16\pi \int_0^3 (3 - x) \,dx & \text{[Integrate with respect to } x \text{]} \\ &= 16\pi \left. \left( 3x - \frac{x^2}{2} \right) \right|_0^3 & \text{[Fundamental Theorem of Calculus]} \\ &= 16\pi \left( \left[ 3(3) - \frac{3^2}{2} \right] - [ 0 ] \right) & \text{[Simplify]} \\ &= 16\pi \left( 9 - \frac{9}{2} \right) \\ &= 16\pi \left( \frac{9}{2} \right) \\ Volume &= \boxed{ 72\pi } & \text{[Volume of Revolution of } R \text{ about the } X \text{-Axis]} \end{align*} {/eq}

B. Volume of Revolution about the Y-Axis using the Shell Method

{eq}\begin{align*} Volume &= 2\pi \int_0^3 x \left( 4 \sqrt{3} - 4 \sqrt{x} \right) \,dx & \text{[The } y \text{-axis is also the vertical line } x = 0 ] \\ &= 2\pi \int_0^3 4 \left( x \sqrt{3} - x^\frac{3}{2} \right) \,dx & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= 8\pi \int_0^3 \left( x \sqrt{3} - x^\frac{3}{2} \right) \,dx & \text{[Integrate with respect to } x \text{]} \\ &= 8\pi \left. \left( \frac{ x^2 \sqrt{3} }{2} - \frac{2}{5} x^\frac{5}{2} \right) \right|_0^3 & \text{[Fundamental Theorem of Calculus]} \\ &= 8\pi \left( \left[ \frac{ 3^2 \sqrt{3} }{2} - \frac{2}{5} (3)^\frac{5}{2} \right] - [ 0 ] \right) & \text{[Simplify]} \\ &= 8\pi \left( \frac{9\sqrt{3}}{2} - \frac{18\sqrt{3}}{5} \right) \\ &= 8\pi \left( \frac{9\sqrt{3}}{10} \right) \\ Volume &= \boxed{ \frac{ 36\pi \sqrt{3} }{5} } & \text{[Volume of Revolution of } R \text{ about the } Y \text{-Axis]} \end{align*} {/eq}

C. Volume of Revolution about the Horizontal Line {eq}y = 4 \sqrt{3} {/eq} using the Disk/Washer Method

{eq}\begin{align*} Volume &= -\pi \int_0^3 \left[ \left( 4 \sqrt{3} - 4 \sqrt{3} \right)^2 - \left( 4 \sqrt{x} - 4 \sqrt{3} \right)^2 \right] \,dx & \text{[The volume is equal to the negative of the integral since } R \text{ is below the axis of revolution]} \\ &= -\pi \int_0^3 \left[ 0 - (4)^2 \left( \sqrt{x} - \sqrt{3} \right)^2 \right] \,dx \\ &= -\pi \int_0^3 -16 \left( x - 2x^\frac{1}{2} \sqrt{3} + 3 \right) \,dx & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= 16\pi \int_0^3 \left( x - 2x^\frac{1}{2} \sqrt{3} + 3 \right) \,dx & \text{[Integrate with respect to } x \text{]} \\ &= 16\pi \left. \left( \frac{x^2}{2} - \frac{4\sqrt{3}}{3} x^\frac{3}{2} + 3x \right) \right|_0^3 & \text{[Fundamental Theorem of Calculus]} \\ &= 16\pi \left( \left[ \frac{3^2}{2} - \frac{4\sqrt{3}}{3} (3)^\frac{3}{2} + 3(3) \right] - [ 0 ] \right) & \text{[Simplify]} \\ &= 16\pi \left( \frac{9}{2} - 12 + 9 \right) \\ &= 16\pi \left( \frac{3}{2} \right) \\ Volume &= \boxed{ 24\pi } & \left[ \text{Volume of Revolution of } R \text{ about the Horizontal Line } y = 4\sqrt{3} \right] \end{align*} {/eq}

D. Volume of Revolution about the Vertical Line {eq}x = 3 {/eq} using the Shell Method

{eq}\begin{align*} Volume &= -2\pi \int_0^3 (x - 3) \left( 4 \sqrt{3} - 4 \sqrt{x} \right) \,dx & \text{[The volume is equal to the negative of the integral since } R \text{ is on the right of the axis of revolution]} \\ &= -2\pi \int_0^3 4 (x - 3) \left( \sqrt{3} - \sqrt{x} \right) \,dx & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= -8\pi \int_0^3 \left( - x^\frac{3}{2} + x \sqrt{3} + 3x^\frac{1}{2} - 3\sqrt{3} \right) \,dx & \text{[Integrate with respect to } x \text{]} \\ &= -8\pi \left. \left( - \frac{2}{5} x^\frac{5}{2} + \frac{x^2 \sqrt{3}}{2} + 2x^\frac{3}{2} - 3x \sqrt{3} \right) \right|_0^3 & \text{[Fundamental Theorem of Calculus]} \\ &= -8\pi \left( \left[ - \frac{2}{5} (3)^\frac{5}{2} + \frac{(3)^2 \sqrt{3}}{2} + 2(3)^\frac{3}{2} - 3(3) \sqrt{3} \right] - [ 0 ] \right) & \text{[Simplify]} \\ &= -8\pi \left( - \frac{18\sqrt{3}}{5} + \frac{9\sqrt{3}}{2} + 6\sqrt{3} - 9 \sqrt{3} \right) \\ &= -8\pi \left( - \frac{21\sqrt{3}}{10} \right) \\ Volume &= \boxed{ \frac{ 84\pi \sqrt{3} }{5} } & \text{[Volume of Revolution of } R \text{ about the Vertical Line } x = 3 ] \end{align*} {/eq}