(1 point) Find the volume of the solid enclosed by the parabolic cylinder 𝑦=𝑥2y=x2 and the planes 𝑧=5+𝑦z=5+y and 𝑧=6𝑦z=6y by subtracting two volumes.

Volume =∫𝑏𝑎∫𝑑𝑐=∫ab∫cd 

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 𝑑𝑦 𝑑𝑥−∫𝑏𝑎∫𝑑𝑐dy dx−∫ab∫cd 

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 𝑑𝑦 𝑑𝑥dy dx

where

𝑎=a= 

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𝑏=b= 

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𝑐=c= 

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𝑑=d= 

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Find the volume.

Volume == 

To find the volume of the solid enclosed by the parabolic cylinder \( y = x^2 \) and the planes \( z = 5 + y \) and \( z = 6y \), we will set up two integrals for the volumes determined by each plane and then subtract one from the other.

### Step-by-Step Solution:

1. **Identify Bounds in the \( x \)- and \( y \)-directions:**
   - The parabolic cylinder \( y = x^2 \) suggests that \( y \) ranges from \( 0 \) to \( 1 \) (based on the overlap of the two planes within this interval).
   - Consequently, \( x \) ranges from \( -1 \) to \( 1 \) because \( x^2 \leq 1 \) over the interval of interest.

2. **Set Up the Double Integral for Each Plane:**
   - **Volume 1** (bounded by \( z = 6y \)):
     \[
     V_1 = \int_{-1}^{1} \int_{0}^{x^2} 6y \, dy \, dx
     \]
   - **Volume 2** (bounded by \( z = 5 + y \)):
     \[
     V_2 = \int_{-1}^{1} \int_{0}^{x^2} (5 + y) \, dy \, dx
     \]

3. **Find the Volume by Subtraction**:
   The volume enclosed by the two surfaces will be given by:
   \[
   V = V_2 - V_1 = \int_{-1}^{1} \int_{0}^{x^2} \big((5 + y) - 6y\big) \, dy \, dx
   \]
   Simplify the integrand:
   \[
   V = \int_{-1}^{1} \int_{0}^{x^2} (5 - 5y) \, dy \, dx
   \]

4. **Evaluate the Inner Integral with Respect to \( y \):**
   \[
   V = \int_{-1}^{1} \left[ 5y - \frac{5y^2}{2} \right]_{0}^{x^2} \, dx
   \]
   Substitute \( y = x^2 \) into the evaluated expression:
   \[
   V = \int_{-1}^{1} \left( 5x^2 - \frac{5(x^2)^2}{2} \right) \, dx
   \]
   Simplify further:
   \[
   V = \int_{-1}^{1} \left( 5x^2 - \frac{5x^4}{2} \right) \, dx
   \]

5. **Evaluate the Outer Integral with Respect to \( x \):**
   Since the integrand is an even function, we can double the integral over \( [0, 1] \):
   \[
   V = 2 \int_{0}^{1} \left( 5x^2 - \frac{5x^4}{2} \right) \, dx
   \]
   Compute each term separately:
   \[
   V = 2 \left( \int_{0}^{1} 5x^2 \, dx - \int_{0}^{1} \frac{5x^4}{2} \, dx \right)
   \]
   Evaluate each integral:
   - \( \int_{0}^{1} 5x^2 \, dx = 5 \cdot \frac{x^3}{3} \Big|_{0}^{1} = \frac{5}{3} \)
   - \( \int_{0}^{1} \frac{5x^4}{2} \, dx = \frac{5}{2} \cdot \frac{x^5}{5} \Big|_{0}^{1} = \frac{5}{10} = \frac{1}{2} \)

6. **Combine the Results:**
   \[
   V = 2 \left( \frac{5}{3} - \frac{1}{2} \right) = 2 \cdot \frac{10 - 3}{6} = 2 \cdot \frac{7}{6} = \frac{7}{3}
   \]

### Final Answer:
\[
\text{Volume} = \frac{7}{3}
\]

Would you like more details, or do you have any questions?

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1. How do we identify the bounds for integration with parabolic cylinders?
2. Why do we set up two separate integrals for this problem?
3. Can volume calculation be done by a single integral in similar problems?
4. What makes an integrand function even, and how does that simplify the work?
5. Why do we need to subtract volumes to find the enclosed region?

**Tip:** Check symmetry in integrals; it can often simplify calculations significantly.

Find the volume of the solid enclosed by the parabolic cylinder y = x^2 and the planes z = 5 + y and z = 6y by subtracting two volumes.

Calculate the volume of a solid bounded by the parabolic cylinder y = x^2 and the planes z = 5 + y and z = 6y using double integrals. This problem illustrates the concept of volumes between surfaces and is suitable for advanced calculus students.

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