Find the volume of the solid generated by revolving the region bounded by x=0, y=10, y = x+2 around x=10 using cylindrical shell and washer methods.

Steps to Solve

1. Find the intersection points of the given equations

First, find where $y = x + 2$ and $y = 10$ intersect to define the limits of integration. Set the equations equal to each other: $x + 2 = 10$ $x = 8$ So, the intersection point is at $x = 8$. Since we are also bounded by $x=0$, our limits of integration will range from 0 to 8.

2. Cylindrical Shell Method Setup

The volume using cylindrical shells around $x = 10$ is given by $$V = 2\pi \int_a^b (10-x)h(x) , dx$$ where $a$ and $b$ are the limits of integration along the x-axis, and $h(x)$ is the height of the shell at a given $x$ value. In this case, $a = 0$, $b = 8$, and $h(x) = 10 - (x + 2) = 8 - x$.

3. Cylindrical Shell Method Integration

Now, evaluate the integral: $$V = 2\pi \int_0^8 (10 - x)(8 - x) , dx$$ $$V = 2\pi \int_0^8 (80 - 10x - 8x + x^2) , dx$$ $$V = 2\pi \int_0^8 (x^2 - 18x + 80) , dx$$ $$V = 2\pi \left[ \frac{x^3}{3} - 9x^2 + 80x \right]_0^8$$ $$V = 2\pi \left[ \frac{8^3}{3} - 9(8^2) + 80(8) \right] - 2\pi \left[ 0 \right]$$ $$V = 2\pi \left[ \frac{512}{3} - 576 + 640 \right]$$ $$V = 2\pi \left[ \frac{512}{3} + 64 \right]$$ $$V = 2\pi \left[ \frac{512 + 192}{3} \right]$$ $$V = 2\pi \left[ \frac{704}{3} \right]$$ $$V = \frac{1408\pi}{3}$$

4. Washer Method Setup

Using the washer method with respect to $y$, the volume is given by $$V = \pi \int_c^d (R(y)^2 - r(y)^2) , dy$$ where $c$ and $d$ are the limits of integration along the y-axis, $R(y)$ is the outer radius, and $r(y)$ is the inner radius. Here, the limits of integration are $c = 2$ and $d = 10$. The outer radius is $R(y) = 10 - 0 = 10$, and the inner radius is $r(y) = 10 - (y - 2) = 12 - y$.

5. Washer Method Integration

Now, evaluate the integral: $$V = \pi \int_2^{10} (10^2 - (12 - y)^2) , dy$$ $$V = \pi \int_2^{10} (100 - (144 - 24y + y^2)) , dy$$ $$V = \pi \int_2^{10} (100 - 144 + 24y - y^2) , dy$$ $$V = \pi \int_2^{10} (-y^2 + 24y - 44) , dy$$ $$V = \pi \left[ -\frac{y^3}{3} + 12y^2 - 44y \right]_2^{10}$$ $$V = \pi \left[ -\frac{10^3}{3} + 12(10^2) - 44(10) \right] - \pi \left[ -\frac{2^3}{3} + 12(2^2) - 44(2) \right]$$ $$V = \pi \left[ -\frac{1000}{3} + 1200 - 440 \right] - \pi \left[ -\frac{8}{3} + 48 - 88 \right]$$ $$V = \pi \left[ -\frac{1000}{3} + 760 \right] - \pi \left[ -\frac{8}{3} - 40 \right]$$ $$V = \pi \left[ -\frac{1000}{3} + 760 + \frac{8}{3} + 40 \right]$$ $$V = \pi \left[ -\frac{992}{3} + 800 \right]$$ $$V = \pi \left[ \frac{-992 + 2400}{3} \right]$$ $$V = \pi \left[ \frac{1408}{3} \right]$$ $$V = \frac{1408\pi}{3}$$