Question 1165589: Find the volume of the solid obtained by rotating the region bounded by
x = y^2 - 2 and y = x about the line y = 3. Use themethod of cylindrical shells.
a. 70.69 cubic units b. 80.69 cubic units c. 90.69 cubic units
d. 100.69 cubic units
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=1
import numpy as np
from scipy.integrate import quad
# The intersection points are found by setting x = y^2 - 2 equal to x = y.
# y^2 - 2 = y
# y^2 - y - 2 = 0
# (y - 2)(y + 1) = 0
# Intersection points are y = -1 and y = 2.
# Outer radius R(y) is the distance from the axis y=3 to the lower function (parabola)
# R(y) = 3 - (y)
# Inner radius r(y) is the distance from the axis y=3 to the upper function (line)
# r(y) = 3 - (y^2 - 2) = 5 - y^2
# Wait, the method specified is CYLINDRICAL SHELLS.
# If rotating about a horizontal line (y=3), the cylindrical shells method uses integration with respect to x.
# The formula for cylindrical shells rotating about a horizontal axis y=k is:
# V = 2 * pi * integral (k - y) * h(y) dy or 2 * pi * integral (y - k) * h(y) dy
# **Correction:** When using the Cylindrical Shells Method, the integral must be with respect to the variable *perpendicular* to the axis of rotation.
# Since the axis of rotation is y = 3 (horizontal), we must integrate with respect to **y**.
# V = 2 * pi * integral (Radius) * (Height) dy
# 1. Radius (h): The distance from a point y to the axis of rotation y=3.
# The region is below the axis y=3, so Radius = 3 - y.
# Radius(y) = 3 - y
# 2. Height (h(y)): The horizontal distance between the two curves.
# Right function (greater x) is the line: x_R = y
# Left function (smaller x) is the parabola: x_L = y^2 - 2
# Height(y) = x_R - x_L = y - (y^2 - 2) = y - y^2 + 2
# 3. Limits of integration: y-coordinates of intersection, from y=-1 to y=2.
# 4. Integrand: 2 * pi * (3 - y) * (y - y^2 + 2)
def integrand(y):
radius = 3 - y
height = y - y**2 + 2
return radius * height
# Expand the integrand (Optional, but good for verification):
# (3 - y)(2 + y - y^2) = 6 + 3y - 3y^2 - 2y - y^2 + y^3
# = y^3 - 4y^2 + y + 6
def expanded_integrand(y):
return y**3 - 4*y**2 + y + 6
# Calculate the definite integral
a_limit = -1
b_limit = 2
integral_value, error = quad(expanded_integrand, a_limit, b_limit)
# Volume V = 2 * pi * integral_value
Volume = 2 * np.pi * integral_value
print(f"Integral Value: {integral_value}")
print(f"Volume: {Volume}")
```
```text?code_stdout&code_event_index=1
Integral Value: 11.249999999999998
Volume: 70.68583470577033
```
This problem requires using the **Cylindrical Shells Method** for a region rotated around a horizontal axis ($y=3$). Since the axis of rotation is horizontal, the integration must be done with respect to $y$.
## 1\. Find the Intersection Points
First, find the $y$-coordinates where the two curves intersect by setting their $x$-expressions equal to each other:
$$y = y^2 - 2$$
$$0 = y^2 - y - 2$$
$$0 = (y - 2)(y + 1)$$
The curves intersect at $\mathbf{y = -1}$ and $\mathbf{y = 2}$. These are the limits of integration.
## 2\. Set up the Shells Integral
The formula for the volume using the Cylindrical Shells Method, rotated about a horizontal line $y=k$, is:
$$V = 2\pi \int_{y_1}^{y_2} (\text{Radius}) \times (\text{Height}) \, dy$$
### a) Radius ($R$)
The radius is the distance from the axis of rotation ($y=3$) to the representative strip at $y$. Since the region is below $y=3$, the radius is:
$$R(y) = 3 - y$$
### b) Height ($H$)
The height of the cylindrical shell is the horizontal distance between the two curves:
$$H(y) = x_{\text{Right}} - x_{\text{Left}}$$
$$H(y) = y - (y^2 - 2) = 2 + y - y^2$$
### c) Volume Integral
$$V = 2\pi \int_{-1}^{2} (3 - y)(2 + y - y^2) \, dy$$
## 3\. Evaluate the Integral
First, expand the integrand:
$$(3 - y)(2 + y - y^2) = 6 + 3y - 3y^2 - 2y - y^2 + y^3$$
$$ = y^3 - 4y^2 + y + 6$$
Now, integrate the polynomial:
$$V = 2\pi \left[ \frac{y^4}{4} - \frac{4y^3}{3} + \frac{y^2}{2} + 6y \right]_{-1}^{2}$$
**Evaluate at $y=2$:**
$$\left( \frac{2^4}{4} - \frac{4(2^3)}{3} + \frac{2^2}{2} + 6(2) \right) = \left( 4 - \frac{32}{3} + 2 + 12 \right) = 18 - \frac{32}{3} = \frac{54 - 32}{3} = \frac{22}{3}$$
**Evaluate at $y=-1$:**
$$\left( \frac{(-1)^4}{4} - \frac{4(-1)^3}{3} + \frac{(-1)^2}{2} + 6(-1) \right) = \left( \frac{1}{4} + \frac{4}{3} + \frac{1}{2} - 6 \right)$$
$$ = \left( \frac{3}{12} + \frac{16}{12} + \frac{6}{12} - \frac{72}{12} \right) = \frac{25 - 72}{12} = -\frac{47}{12}$$
**Subtract the values:**
$$\text{Integral Value} = \frac{22}{3} - \left(-\frac{47}{12}\right) = \frac{22}{3} + \frac{47}{12} = \frac{88}{12} + \frac{47}{12} = \frac{135}{12}$$
$$\text{Integral Value} = \frac{45}{4} = 11.25$$
**Final Volume:**
$$V = 2\pi \times 11.25 = 22.5\pi$$
$$V \approx 70.6858$$
The volume is approximately **$70.69$ cubic units**.
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The correct option is **a. 70.69 cubic units**.
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