Let $R$ be the region enclosed by the curves $y=\sqrt{x}$ and $\newline$ $y=\frac{x}{3} \text {. }$ $\newline$ A solid is generated by rotating $R$ about the line $x=-1$ . $\newline$ What is the volume of the solid? $\newline$ Give an exact answer in terms of $\pi$ . $\newline$ $\square$

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Write and solve direct variation equations

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Q. Let

R

be the region enclosed by the curves

y=\sqrt{x}

and

\newline

y=\frac{x}{3} \text {. }

\newline

A solid is generated by rotating

R

about the line

x=-1

\newline

What is the volume of the solid?

\newline

Give an exact answer in terms of

\pi

\newline

\square

Find Intersection Points: First, we need to find the intersection points of $y=\sqrt{x}$ and $y=\frac{x}{3}$ to determine the limits of integration. $\newline$ Set $\sqrt{x} = \frac{x}{3}$ and solve for $x$ . $\newline$ $x = \left(\frac{x}{3}\right)^2$ $\newline$ $x = \frac{x^2}{9}$ $\newline$ $9x = x^2$ $\newline$ $x^2 - 9x = 0$ $\newline$ $x(x - 9) = 0$ $\newline$ $x = 0$ or $y=\frac{x}{3}$ $0$ $\newline$ So, the intersection points are at $y=\frac{x}{3}$ $1$ and $y=\frac{x}{3}$ $2$ .
Set Up Integral: Now, we'll set up the integral for the volume using the washer method since we're rotating around a line that is not the y-axis. $\newline$ The volume $V$ is given by the integral from $0$ to $9$ of $\pi(\text{outer radius}^2 - \text{inner radius}^2) \, dx$ .
Determine Outer and Inner Radii: The outer radius is the distance from the line $x=-1$ to the curve $y=\sqrt{x}$ , which is $(x - (-1)) = x + 1$ . The inner radius is the distance from the line $x=-1$ to the curve $y=\frac{x}{3}$ , which is $(x - (-1)) = x + 1$ . But we need to express these in terms of $y$ , so we solve for $x$ in terms of $y$ . For $y=\sqrt{x}$ , $y=\sqrt{x}$ $0$ . For $y=\frac{x}{3}$ , $y=\sqrt{x}$ $2$ .
Change Limits of Integration: Now we substitute $x$ with $y^2$ and $3y$ into the outer and inner radius expressions, respectively. $\newline$ Outer radius: $(y^2 + 1)$ $\newline$ Inner radius: $(3y + 1)$
Write Integral for Volume: We also need to change the limits of integration from $x$ to $y$ . When $x=0$ , $y=0$ (for both curves). When $x=9$ , $y=3$ (for $y=\sqrt{x}$ ) and $y=3$ (for $y=\frac{x}{3}$ ). So the limits of integration in terms of $y$ are from $y$ $0$ to $y$ $1$ .
Expand and Subtract Integrand: Now we can write the integral for the volume $V$ . $V = \pi \int_{0}^{3} [(y^2 + 1)^2 - (3y + 1)^2] \, dy$ .
Expand and Subtract Integrand: Now we can write the integral for the volume $V$ . $V = \pi \cdot \int_{0}^{3} [(y^2 + 1)^2 - (3y + 1)^2] \, dy$ .Let's expand the integrand before integrating. $(y^2 + 1)^2 = y^4 + 2y^2 + 1$ $(3y + 1)^2 = 9y^2 + 6y + 1$ Now subtract the inner radius squared from the outer radius squared. $(y^4 + 2y^2 + 1) - (9y^2 + 6y + 1) = y^4 - 7y^2 - 6y$