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# 数学代写|微积分代写Calculus代考|MATH-172 THE METHOD OF WASHERS

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## 数学代写|微积分代写Calculus代考|THE METHOD OF WASHERS

EXAMPLE 8.5
A solid is formed by rotating the triangle with vertices $(0,0),(2,0)$, and $(1,1)$ about the $x$-axis. See Fig. 8.15. What is the resulting volume?

For $0 \leq x \leq 1$, the upper edge of the triangle has equation $y=x$. Thus the segment being rotated extends from $(x, 0)$ to $(x, x)$. Under rotation, it will generate a disk of radius $x$, and hence area $A(x)=\pi x^2$. Thus the volume generated over the segment $0 \leq x \leq 1$ is
$$V_1=\int_0^1 \pi x^2 d x .$$
Similarly, for $1 \leq x \leq 2$, the upper edge of the triangle has equation $y=$ $2-x$. Thus the segment being rotated extends from $(x, 0)$ to $(x, 2-x)$. Under rotation, it will generate a disk of radius $2-x$, and hence area $A(x)=\pi(2-x)^2$. Thus the volume generated over the segment $1 \leq x \leq 2$ is
$$V_2=\int_1^2 \pi(2-x)^2 d x$$

## 数学代写|微积分代写Calculus代考|THE METHOD OF CYLINDRICAL SHELLS

Our philosophy will now change. When we divide our region up into vertical strips, we will now rotate each strip about the $y$-axis instead of the $x$-axis. Thus, instead of generating a disk with each strip, we will now generate a cylinder.

Look at Fig. 8.19. When a strip of height $h$ and thickness $\Delta x$, with distance $r$ from the $y$-axis, is rotated about the $y$-axis, the resulting cylinder has surface area $2 \pi r \cdot h$ and volume about $2 \pi r \cdot h \cdot \Delta x$. This is the expression that we will treat in order to sum up the volumes of the cylinders.

Use the method of cylindrical shells to calculate the volume of the solid enclosed when the curve $y=x^2, 1 \leq x \leq 3$, is rotated about the $y$-axis.
SOLUTION
As usual, we think of the region under $y=x^2$ and above the $x$-axis as composed of vertical segments or strips. The segment at position $x$ has height $x^2$. Thus, in this instance, $h=x^2, r=x$, and the volume of the cylinder is $2 \pi x \cdot x^2 \cdot \Delta x$. As a result, the requested volume is
$$V=\int_1^3 2 \pi x \cdot x^2 d x$$
We easily calculate this to equal
$$V=2 \pi \cdot \int_1^3 x^3 d x=\left.2 \pi \frac{x^4}{4}\right|_1 ^3=2 \pi\left[\frac{3^4}{4}-\frac{1^4}{4}\right]=40 \pi .$$

## 数学代写|微积分代写Calculus代考|THE METHOD OF WASHERS

$$V_1=\int_0^1 \pi x^2 d x$$

$$V_2=\int_1^2 \pi(2-x)^2 d x$$

## 数学代写|微积分代写Calculus代考|THE METHOD OF CYLINDRICAL SHELLS

$$V=\int_1^3 2 \pi x \cdot x^2 d x$$

$$V=2 \pi \cdot \int_1^3 x^3 d x=\left.2 \pi \frac{x^4}{4}\right|_1 ^3=2 \pi\left[\frac{3^4}{4}-\frac{1^4}{4}\right]=40 \pi .$$

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