How do you find the volume of the region enclosed by the curves $y = 2 x$ $y=2x$ , $y = x^{2}$ $y=x^2$ rotated about the x-axis?

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How do you find the volume of the region enclosed by the curves $y = 2 x$ $y=2x$ , $y = x^{2}$ $y=x^2$ rotated about the x-axis?

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Answer 1 · 2015-09-11T03:44:45

The volume of the solid generated by $y = 2 x$ $y = 2x$ , $y = x^{2}$ $y = x^2$ revolved about the x-axis is $\frac{64 π}{15}$ $(64pi)/15$ .

Explanation:

Graph of the two curves

Revolving the area between these two curves about the x-axis, we end up with something that looks sort of like a cone... a hollow cone, with a curved inside. Now, imagine for a second taking a cross section parallel to the y-z plane, cutting the cone down the middle. The cross-section is going to look like a washer, with an inner radius equal to the height of $y = x^{2}$ $y = x^2$ wherever we are in the solid, and the outer radius equal to $y = 2 x$ $y = 2x$ at that same $x$ $x$ .

For reference, the area of a washer is equal to $π \cdot {(r_{outer})}_{outer}^{2} - π \cdot {(r_{inner})}_{inner}^{2}$ $pi*(r_"outer")^2 - pi*(r_"inner")^2$ , if $r_{outer}$ $r_"outer"$ is the outer radius and $r_{inner}$ $r_"inner"$ is the inner radius. This fact should be immediately obvious - we're just subtracting the area of a circle from the area of another circle.

Now, to find the volume of the solid, we need to sum (integrate) the area of each of cross-sectional washer in the solid. This procedure is commonly called the method of washers .

So, we'll use the method of washers to find the volume of this solid.

The general formula is

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

where $f (x)$ $f(x)$ is a function giving the outer radius of the washer at any x, and $g (x)$ $g(x)$ is a function giving the inner radius of the washer. Note that the integrands here represent the areas of circles - and together, they give the area of each infinitesimal washer.

In our case, $f (x)$ $f(x)$ is equal to $2 x$ $2x$ and $g (x)$ $g(x)$ is equal to $x^{2}$ $x^2$ .

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

Next, we need to worry about the limits of the integrals. Since we're only concerned with rotating the region between $x^{2}$ $x^2$ and $2 x$ $2x$ , the limits should involve the locations where $y = x^{2}$ $y = x^2$ and $y = 2 x$ $y = 2x$ intersect. Intersection occurs at $x = 0$ $x = 0$ and $x = 2$ $x = 2$ .

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

The difficult part - setting up the appropriate integral with correct limits - is done, so I won't walk through evaluating the integral. It evaluates to $\frac{64 π}{15}$ $(64pi)/15$ , but if you don't trust me you can evaluate it yourself as an exercise.

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How do you find the volume of the region enclosed by the curves $y = 2 x$ $y=2x$ , $y = x^{2}$ $y=x^2$ rotated about the x-axis?

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How do you find the volume of the region enclosed by the curves y=2xy=2x, y=x2y=x^2 rotated about the x-axis?

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Explanation:

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How do you find the volume of the region enclosed by the curves $y = 2 x$ $y=2x$ , $y = x^{2}$ $y=x^2$ rotated about the x-axis?