Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. y =x', y = x; about y=-3

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Chapter1: Functions And Models
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### Volume of Solid Rotation

To find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, follow the steps below.

Given curves:
\[ y = x^2 \]
\[ y^2 = x \]

Axis of rotation:
\[ y = -3 \]

#### Step-by-Step Solution:

1. **Identify the Points of Intersection:**
   Find the points where the curves \( y = x^2 \) and \( y^2 = x \) intersect.

   - For \( y = x^2 \): \( x = y^2 \);
   - For \( y^2 = x \): \( x = y^2 \).

  Equate the two expressions for \( x \):
  \[
  y^2 = y^4
  \]

  Solving this equation gives:
  \[
  y = 0 \quad \text{or} \quad y = 1
  \]

2. **Set Up the Integral:**
   Using the disk/washer method, set up the integral for the volume.

   The outer radius (R) is the distance from the line \( y = -3 \) to the curve \( y = x^2 \):
   \[
   R = x^2 + 3
   \]

   The inner radius (r) is the distance from the line \( y = -3 \) to the curve \( y = \sqrt{x} \):
   \[
   r = \sqrt{x} + 3
   \]

3. **Volume Integral:**
   The volume of the solid of revolution about \( y = -3 \) is given by the integral:
   \[
   V = \pi \int_{0}^{1} [(R)^2 - (r)^2] \, dx
   \]
   Where \( R = x^2 + 3 \) and \( r = \sqrt{x} + 3 \).

4. **Evaluate the Integral:**
   Simplify the integral and evaluate it step-by-step:
   \[
   V = \pi \int_{0}^{1} [(x^2 + 3)^2 - (\sqrt{x} + 3)^2] \, dx
   \]

Following these steps will give you the volume of the solid obtained by rotating the specified region around the line \( y = -3 \). 

This
Transcribed Image Text:### Volume of Solid Rotation To find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line, follow the steps below. Given curves: \[ y = x^2 \] \[ y^2 = x \] Axis of rotation: \[ y = -3 \] #### Step-by-Step Solution: 1. **Identify the Points of Intersection:** Find the points where the curves \( y = x^2 \) and \( y^2 = x \) intersect. - For \( y = x^2 \): \( x = y^2 \); - For \( y^2 = x \): \( x = y^2 \). Equate the two expressions for \( x \): \[ y^2 = y^4 \] Solving this equation gives: \[ y = 0 \quad \text{or} \quad y = 1 \] 2. **Set Up the Integral:** Using the disk/washer method, set up the integral for the volume. The outer radius (R) is the distance from the line \( y = -3 \) to the curve \( y = x^2 \): \[ R = x^2 + 3 \] The inner radius (r) is the distance from the line \( y = -3 \) to the curve \( y = \sqrt{x} \): \[ r = \sqrt{x} + 3 \] 3. **Volume Integral:** The volume of the solid of revolution about \( y = -3 \) is given by the integral: \[ V = \pi \int_{0}^{1} [(R)^2 - (r)^2] \, dx \] Where \( R = x^2 + 3 \) and \( r = \sqrt{x} + 3 \). 4. **Evaluate the Integral:** Simplify the integral and evaluate it step-by-step: \[ V = \pi \int_{0}^{1} [(x^2 + 3)^2 - (\sqrt{x} + 3)^2] \, dx \] Following these steps will give you the volume of the solid obtained by rotating the specified region around the line \( y = -3 \). This
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