Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves $x = 0$ $x=0$ , $y = 0$ $y=0$ and $y = - 2 x + 2$ $y=-2x+2$ ?

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Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves $x = 0$ $x=0$ , $y = 0$ $y=0$ and $y = - 2 x + 2$ $y=-2x+2$ ?

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Answer 1 · 2015-09-11T21:18:47

$V = \frac{4 π}{3}$ $V = (4pi)/3$

Explanation:

Plot of y = 2 - 2x

Looking at the graph of $y = 2 - 2 x$ $y = 2 - 2x$ , imagine the area between the line, the x-axis, and the y-axis, being revolved around the x-axis. You'll end up with a cone, with the point/tip at $x = 1$ $x = 1$ , and the center of the circular base (which has radius 2) at the origin.

Next, imagine looking at a cross-section parallel to the x-z axis - parallel to the circular base. Every cross-section will be a circle. At $x = 0$ $x = 0$ , the circle has radius 2, and at $x = 1$ $x = 1$ , the circle has radius 0. In fact, for any $x$ $x$ , the cross-sectional circle has radius $2 - 2 x$ $2 - 2x$ .

To find the volume of the solid of revolution, we can imagine that our solid is composed of infinitely many disks, of infinitesimal width, and radius equal to $2 - 2 x$ $2 - 2x$ . To find the volume of the solid, we sum up (integrate) each disk. This process is commonly called the method of disks.

The general formula for the method of disks is:

$V = \int_{a}^{b} π \cdot {(f (x))}^{2} d x$ $V = int_a^b pi*(f(x))^2 dx$

where $f (x)$ $f(x)$ is the curve we're revolving about the x-axis, and $[a, b]$ $[a,b]$ is the interval we're concerned with. Important to note is the $π \cdot {(f (x))}^{2}$ $pi*(f(x))^2$ - this just means "area of a circle with radius $f (x)$ $f(x)$ ." Multiplying this by $d x$ $dx$ gives the volume of a cylinder (disk) of width $d x$ $dx$ and radius $d x$ $dx$ . And integration allows us to sum all these disks up, exactly what we want to accomplish.

So, in our case, we'll note that $f (x) = 2 - 2 x$ $f(x) = 2 - 2x$ , and $[a, b] = [0, 1]$ $[a,b] = [0,1]$ , and substitute:

$V = \int_{0}^{1} π \cdot {(2 - 2 x)}^{2} d x$ $V = int_0^1 pi*(2 - 2x)^2 dx$

Well, that was the difficult part; setting up the integral. From here, evaluating the integral should be fairly easy. I'll leave it to you as an exercise, but the answer should come out to

$V = \frac{4 π}{3}$ $V = (4pi)/3$

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Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves $x = 0$ $x=0$ , $y = 0$ $y=0$ and $y = - 2 x + 2$ $y=-2x+2$ ?

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Explanation:

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Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves x=0x=0, y=0y=0 and y=−2x+2y=-2x+2?

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Using the disk method, how do you find the volume of the solid generated by revolving about the x-axis the area bounded by the curves $x = 0$ $x=0$ , $y = 0$ $y=0$ and $y = - 2 x + 2$ $y=-2x+2$ ?