Respuesta :
Answer:
In previous applications of integration, we required the function f(x) to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for f(x). Here, we require f(x) to be differentiable, and furthermore we require its derivative, f′(x), to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)
Let f(x) be a smooth function defined over [a,b]. We want to calculate the length of the curve from the point (a,f(a)) to the point (b,f(b)). We start by using line segments to approximate the length of the curve. For i=0,1,2,…,n, let P={xi} be a regular partition of [a,b]. Then, for i=1,2,…,n, construct a line segment from the point (xi−1,f(xi−1)) to the point (xi,f(xi)). Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. Figure 2.37 depicts this construct for n=5.
This figure is a graph in the first quadrant. The curve increases and decreases. It is divided into parts at the points a=xsub0, xsub1, xsub2, xsub3, xsub4, and xsub5=b. Also, there are line segments between the points on the curve.
Figure 2.37 We can approximate the length of a curve by adding line segments.
To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by Δx. The change in vertical distance varies from interval to interval, though, so we use Δyi=f(xi)−f(xi−1) to represent the change in vertical distance over the interval [xi−1,xi], as shown in Figure 2.38. Note that some (or all) Δyi may be negative.
This figure is a graph. It is a curve above the x-axis beginning at the point f(xsubi-1). The curve ends in the first quadrant at the point f(xsubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.
Figure 2.38 A representative line segment approximates the curve over the interval [xi−1,xi].
By the Pythagorean theorem, the length of the line segment is
√
(Δx)2+(Δyi)2
. We can also write this as Δx
√
1+((Δyi)/(Δx))2
. Now, by the Mean Value Theorem, there is a point x
*
i
∈[xi−1,xi] such that f′(x
*
i
)=(Δyi)/(Δx). Then the length of the line segment is given by Δx
√
1+[f′(x
*
i
)]2
. Adding up the lengths of all the line segments, we get
Arc Length≈
n
∑
i=1
√
1+[f′(x
*
i
)]2
Δx.
This is a Riemann sum. Taking the limit as n→∞, we have
Step-by-step explanation:
The definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the [tex]x-axis[/tex], [tex]y = 81 - x^2[/tex], [tex]-5 \leq x \leq 5[/tex] will be [tex]A=\frac{-4360\pi }{3}[/tex].
What are definite integral ?
Definite integral is the area under a curve between two fixed limits. i.e.
[tex]A=2\pi \int\limits^a_b {f(x)}\ dx[/tex]
Here, [tex]a[/tex] and [tex]b[/tex] are limit of [tex]x[/tex].
We have,
[tex]y = 81 - x^2[/tex] and [tex]-5 \leq x \leq 5[/tex]
And,
[tex]f(x)=y = 81 - x^2[/tex]
So, Using above given formula to find the area;
[tex]A=2\pi \int\limits^a_b {f(x)} \ dx[/tex]
So,
[tex]A=2\pi \int\limits^{-5}_5 ({81-x^2}) \ dx[/tex]
[tex]A=2\pi \int\limits^{-5}_5 ({81x-\frac{x^3}{3} }) \ dx[/tex]
[tex]A=\frac{2\pi}{3} \int\limits^{-5}_5 ({243x-x^3) \ dx[/tex]
[tex]A= \frac{2\pi }{3} \left| 243x-x^3 \right| \limits^{-5}_5[/tex]
[tex]A=\frac{2\pi }{3} \left| 243x-x^3 \right| \limits^{-5}_5[/tex]
[tex]A= \frac{2\pi }{3}(243*(-5)-(-5)^3 ) -(243*5-5^3)[/tex]
[tex]A=\frac{2\pi }{3} (-1215+125) -(1215-125)[/tex]
[tex]A=\frac{2\pi }{3} (-1090) -(1090)[/tex]
[tex]A=\frac{2\pi }{3} * (-2180)[/tex]
[tex]A=\frac{-4360\pi }{3}[/tex]
So, this is the area of the surface generated by revolving the curve .
Hence, we can say that the definite integral that represents the area of the surface generated by revolving the curve on the indicated interval about the [tex]x-axis[/tex], [tex]y = 81 - x^2[/tex], [tex]-5 \leq x \leq 5[/tex] will be [tex]A=\frac{-4360\pi }{3}[/tex].
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