Questions: What is the volume of the solid of revolution generated by revolving the area bounded by y=x, y=-x+4, and y=0 about the y-axis? (A) 4 pi / 5 units ^3 (B) 10 pi / 5 units ^3 (C) 2 pi units ^3 (D) 16 pi units ^3

What is the volume of the solid of revolution generated by revolving the area bounded by y=x, y=-x+4, and y=0 about the y-axis?
(A) 4 pi / 5 units ^3
(B) 10 pi / 5 units ^3
(C) 2 pi units ^3
(D) 16 pi units ^3

Transcript text: What is the volume of the solid of revolution generated by revolving the area bounded by $\boldsymbol{y}=\boldsymbol{x}$, $y=-x+4$, and $y=0$ about the $y$-axis? (A) $4 p / 5$ units ${ }^{3}$ (B) $10 p / 5$ units $^{3}$ (C) $2 p$ units ${ }^{3}$ (D) $16 p$ units ${ }^{3}$

Solution

Solution Steps

To find the volume of the solid of revolution generated by revolving the area bounded by $ y = x $, $ y = -x + 4 $, and $ y = 0 $ about the $ y $-axis, we can use the method of cylindrical shells. The volume $ V $ is given by the integral:

\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]

where $ f(x) $ is the height of the shell and $ x $ is the radius. We need to determine the points of intersection to find the limits of integration.

Step 1: Identify the Bounded Region

First, we need to identify the region bounded by the curves $ y = x $, $ y = -x + 4 $, and $ y = 0 $.

The line $ y = x $ intersects the x-axis at $ (0, 0) $.
The line $ y = -x + 4 $ intersects the x-axis at $ (4, 0) $.
The intersection of $ y = x $ and $ y = -x + 4 $ can be found by setting $ x = -x + 4 $: \[ x + x = 4 \implies 2x = 4 \implies x = 2 \] Substituting $ x = 2 $ into $ y = x $, we get $ y = 2 $.

Thus, the region is bounded by the points $ (0, 0) $, $ (2, 2) $, and $ (4, 0) $.

Step 2: Set Up the Integral for Volume

To find the volume of the solid of revolution about the y-axis, we use the method of cylindrical shells. The volume $ V $ is given by: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where $ f(x) $ is the height of the shell and $ x $ is the radius.

Step 3: Determine the Limits of Integration and Functions

The region is divided into two parts:

From $ x = 0 $ to $ x = 2 $, the height is given by $ y = x $.
From $ x = 2 $ to $ x = 4 $, the height is given by $ y = -x + 4 $.

Step 4: Calculate the Volume for Each Part

For $ 0 \leq x \leq 2 $: \[ V_1 = 2\pi \int_{0}^{2} x \cdot x \, dx = 2\pi \int_{0}^{2} x^2 \, dx \] \[ V_1 = 2\pi \left[ \frac{x^3}{3} \right]_{0}^{2} = 2\pi \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = 2\pi \left( \frac{8}{3} \right) = \frac{16\pi}{3} \]
For $ 2 \leq x \leq 4 $: \[ V_2 = 2\pi \int_{2}^{4} x \cdot (-x + 4) \, dx = 2\pi \int_{2}^{4} (4x - x^2) \, dx \] \[ V_2 = 2\pi \left[ 2x^2 - \frac{x^3}{3} \right]_{2}^{4} \] \[ V_2 = 2\pi \left( \left( 2(4)^2 - \frac{(4)^3}{3} \right) - \left( 2(2)^2 - \frac{(2)^3}{3} \right) \right) \] \[ V_2 = 2\pi \left( \left( 32 - \frac{64}{3} \right) - \left( 8 - \frac{8}{3} \right) \right) \] \[ V_2 = 2\pi \left( \frac{96}{3} - \frac{64}{3} - \frac{24}{3} + \frac{8}{3} \right) = 2\pi \left( \frac{16}{3} \right) = \frac{32\pi}{3} \]

Step 5: Sum the Volumes

The total volume $ V $ is the sum of $ V_1 $ and $ V_2 $: \[ V = V_1 + V_2 = \frac{16\pi}{3} + \frac{32\pi}{3} = \frac{48\pi}{3} = 16\pi \]