(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 can set up an integral that calculates the volume of the region by subtracting two separate integrals, each representing a different volume.

### Step 1: Setting up the volume integral
The volume can be found using the double integral
\[
\text{Volume} = \int_a^b \int_c^d \big((6y) - (5 + y)\big) \, dy \, dx
\]
where:
- \( 6y \) is the upper bound in the \( z \)-direction,
- \( 5 + y \) is the lower bound in the \( z \)-direction.

### Step 2: Determining the bounds
1. The region is bounded by \( y = x^2 \) in the \( xy \)-plane.
2. Since \( y = x^2 \), \( y \) ranges from \( 0 \) to \( 1 \) (assuming the bounds for \( y \) are determined by the intersection in the context).
3. Correspondingly, \( x \) would range from \( -1 \) to \( 1 \).

Thus, we can set:
- \( a = -1 \),
- \( b = 1 \),
- \( c = 0 \),
- \( d = x^2 \).

### Step 3: Setting up the integral for the volume
Now the volume integral becomes:
\[
\text{Volume} = \int_{-1}^1 \int_0^{x^2} \big((6y) - (5 + y)\big) \, dy \, dx.
\]

### Step 4: Simplify the integrand
The integrand \( (6y - (5 + y)) \) simplifies to:
\[
6y - 5 - y = 5y - 5.
\]
Thus, the integral becomes:
\[
\text{Volume} = \int_{-1}^1 \int_0^{x^2} (5y - 5) \, dy \, dx.
\]

### Step 5: Evaluate the inner integral
Now we integrate with respect to \( y \):
\[
\int_0^{x^2} (5y - 5) \, dy = \left[ \frac{5y^2}{2} - 5y \right]_0^{x^2} = \frac{5(x^2)^2}{2} - 5(x^2).
\]
Simplifying further, we get:
\[
= \frac{5x^4}{2} - 5x^2.
\]

### Step 6: Evaluate the outer integral
Now we integrate with respect to \( x \):
\[
\text{Volume} = \int_{-1}^1 \left( \frac{5x^4}{2} - 5x^2 \right) dx.
\]

Splitting this into two separate integrals:
\[
\text{Volume} = \int_{-1}^1 \frac{5x^4}{2} \, dx - \int_{-1}^1 5x^2 \, dx.
\]

### Step 7: Evaluate each integral
1. **First integral**:
   \[
   \int_{-1}^1 \frac{5x^4}{2} \, dx = \frac{5}{2} \int_{-1}^1 x^4 \, dx = \frac{5}{2} \cdot \frac{2}{5} = 0.


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.

Learn to calculate the volume of a solid bounded by the parabolic cylinder y = x^2 and planes z = 5 + y and z = 6y using double integration. This advanced calculus problem involves setting up and simplifying integrals to find the enclosed volume, suitable for college-level students.

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