Question

Найдите объём тела, образованного вращением около оси абсцисс фигуры, ограниченной прямой y = 8x и графиком функции y = 2x^3 при x ≥ 0. Выполните рисунок.

Answer 1

Ответ:

$V=\dfrac{2048\pi }{21}$

Объяснение:

Объем тела, образованного вращением около оси абсцисс фигуры, ограниченной графиком функции y = f(x), находится по формуле:

$V=\int\limits^a_b {\pi f^2(x)} \, dx$

Чтобы найти объем тела, ограниченного графиками двух функций, надо от объема тела, ограниченного верхним графиком, отнять объем тела, ограниченного нижним графиком.

Графики функций y = 8x и  y = 2x³ на рисунке.

Найдем точки пересечения графиков (пределы интегрирования):

$2x^3=8x$

$2x^3-8x=0$

$x(x^2-4)=0$

$x(x-2)(x+2)=0$

$x=0,\; \: x=2,\; \: x=-2$

Так как по условию х ≥ 0, то пределы интегрирования от 0 до 2.

$V=\int\limits^2_0 {\pi (8x)^2} \, dx -\int\limits^2_0 {\pi (2x^3)^2} \, dx =$

$=\int\limits^2_0 {\pi 64x^2} \, dx -\int\limits^2_0 {\pi 4x^6} \, dx =$

$=64\pi \cdot \dfrac{x^3}{3}\bigg|\limits^2_0-4\pi \cdot \dfrac{x^7}{7}\bigg|\limits^2_0=\pi \cdot \dfrac{2^6\cdot 2^3}{3}-\pi\cdot \dfrac{2^2\cdot 2^7}{7}=$

$=2^9\cdot \pi \cdot \left(\dfrac{1}{3}-\dfrac{1}{7}\right)=\dfrac{2^9\cdot \pi\cdot 4}{21}=\dfrac{2048\pi}{21}$

$\boldsymbol{V=\dfrac{2048\pi }{21}}$

Answer 2

Calculation of the Volume of the Solid of Revolution

To find the volume of the solid formed by rotating the figure bounded by the line y = 8x and the graph of the function y = 2x^3 around the x-axis, we can use the method of cylindrical shells.

The volume of the solid can be calculated using the formula:

V = 2π ∫[a,b] x * f(x) dx

where a and b are the x-values where the two curves intersect.

To find the intersection points, we can set the two equations equal to each other:

8x = 2x^3

Simplifying the equation:

2x^3 - 8x = 0

Factoring out 2x:

2x(x^2 - 4) = 0

This equation has three solutions: x = 0, x = -2, and x = 2. However, since we are only interested in the region where x ≥ 0, we can discard the negative solution.

Therefore, the intersection points are x = 0 and x = 2.

Now we can calculate the volume using the formula mentioned earlier. We integrate from x = 0 to x = 2:

V = 2π ∫[0,2] x * (8x - 2x^3) dx

Simplifying the integrand:

V = 2π ∫[0,2] (8x^2 - 2x^4) dx

Integrating term by term:

V = 2π [ (8/3)x^3 - (2/5)x^5 ] evaluated from 0 to 2

Evaluating the definite integral:

V = 2π [ (8/3)(2)^3 - (2/5)(2)^5 - (8/3)(0)^3 + (2/5)(0)^5 ]

V = 2π [ (8/3)(8) - (2/5)(32) ]

V = 2π [ (64/3) - (64/5) ]

V = 2π [ (320/15) - (192/15) ]

V = 2π [ 128/15 ]

V = (256π/15)

Therefore, the volume of the solid of revolution is 256π/15 cubic units.

Graphical Representation

To visualize the solid of revolution, we can plot the two curves and shade the region between them. Here is a rough sketch of the graph:

import matplotlib.pyplot as plt
import numpy as np

x = np.linspace(0, 2, 100)
y1 = 8 * x
y2 = 2 * x**3

plt.plot(x, y1, label='y = 8x')
plt.plot(x, y2, label='y = 2x^3')
plt.fill_between(x, y1, y2, where=(y1 > y2), color='gray', alpha=0.5)

plt.xlabel('x')
plt.ylabel('y')
plt.title('Graph of y = 8x and y = 2x^3')
plt.legend()
plt.grid(True)
plt.show()

This code will generate a plot showing the two curves and the shaded region between them.

Please note that the code provided is a basic example and may require additional adjustments to produce an accurate representation of the graph.