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Let the region
R be the area enclosed by the function
f(x)=sqrtx+2, the horizontal line
y=-1 and the vertical lines
x=0 and
x=6. Find the volume of the solid generated when the region
R is revolved about the line
y=6. You may use a calculator and round to the nearest thousandth.

Let the region $\mathrm{R}$ be the area enclosed by the function $f(x)=\sqrt{x}+2$ , the horizontal line $y=-1$ and the vertical lines $x=0$ and $x=6$ . Find the volume of the solid generated when the region $\mathrm{R}$ is revolved about the line $y=6$ . You may use a calculator and round to the nearest thousandth.

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Volume of cubes and rectangular prisms: word problems

Full solution

Q. Let the region

\mathrm{R}

be the area enclosed by the function

f(x)=\sqrt{x}+2

, the horizontal line

y=-1

and the vertical lines

x=0

and

x=6

. Find the volume of the solid generated when the region

\mathrm{R}

is revolved about the line

y=6

. You may use a calculator and round to the nearest thousandth.

Set up integral: First, we need to set up the integral for the volume using the disk method since we're revolving around a horizontal line.
Adjust functions: The volume $V$ of a solid of revolution generated by revolving a region bounded by $y=f(x)$ , $y=g(x)$ , $x=a$ , and $x=b$ around the x-axis is given by the integral from $a$ to $b$ of $\pi(f(x)^2 - g(x)^2) \, dx$ . Here, we're revolving around $y=6$ , so we need to adjust the functions to account for this.
Calculate limits: The adjusted functions will be $(6 - (\sqrt{x} + 2))^2$ for the top function and $(6 - (-1))^2$ for the bottom function. The limits of integration are from $x=0$ to $x=6$ .
Simplify integral: Set up the integral: $V = \pi\int_{0}^{6} [(6 - (\sqrt{x} + 2))^2 - (6 - (-1))^2] \, dx$ .
Calculate integral: Simplify the integral: $V = \pi\int_{0}^{6} [(4 - \sqrt{x})^2 - 49] \,dx$ .
Further simplify: Now we calculate the integral using a calculator: $V \approx \pi\int_{0}^{6} [(16 - 8\sqrt{x} + x) - 49] \,dx$ .
Integrate terms: Further simplify the integral: $V \approx \pi\int_{0}^{6} [-33 - 8\sqrt{x} + x] dx$ .
Plug in limits: Integrate term by term: $V \approx \pi[-33x - \frac{16}{3}x^{\frac{3}{2}} + \frac{1}{2}x^2]$ from $0$ to $6$ .
Calculate values: Plug in the limits of integration: V \approx \pi[\(-33( $6$ ) - ( $16$ / $3$ )( $6$ )^{ $3$ / $2$ } + ( $1$ / $2$ )( $6$ )^ $2$ ] - \pi[ $-33$ ( $0$ ) - ( $16$ / $3$ )( $0$ )^{ $3$ / $2$ } + ( $1$ / $2$ )( $0$ )^ $2$ ].
Simplify square root: Calculate the values: $V \approx \pi[-198 - (\frac{16}{3})(216)^{\frac{1}{2}} + 18]$ .
Finish calculation: Simplify the square root and continue the calculation: $V \approx \pi[-198 - (\frac{16}{3})\cdot14.7 + 18]$ .
Add numbers: Finish the calculation: $V \approx \pi[-198 - 78.4 + 18]$ .
Add numbers: Finish the calculation: $V \approx \pi[-198 - 78.4 + 18]$ .Add up the numbers: $V \approx \pi[-258.4]$ .

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