Use a triple integral to find the volume of the wedge bounded by the parabolic cylinder yequalsxsquared and the planes zequals18minusy and zequals0.

To find the volume of the wedge bounded by the parabolic cylinder \( y = x^2 \) and the planes \( z = 18 - y \) and \( z = 0 \), we can set up a triple integral.

### Step 1: Determine the region of integration
1. **The plane \( z = 0 \)** is the bottom boundary of the wedge.
2. **The plane \( z = 18 - y \)** is the top boundary of the wedge.
3. **The parabolic cylinder \( y = x^2 \)** gives us the relationship between \( x \) and \( y \).

### Step 2: Find the bounds for \( z \), \( y \), and \( x \)
- \( z \) varies from the bottom plane to the top plane:
  \[
  0 \leq z \leq 18 - y
  \]
- \( y \) varies from the parabola \( y = x^2 \) to the point where the plane \( z = 18 - y \) intersects the \( y \)-axis (i.e., the wedge's shadow in the \( xy \)-plane):
  \[
  x^2 \leq y \leq 18
  \]
- \( x \) varies from the minimum \( x \) value for the parabola to the maximum \( x \) value for the parabola:
  \[
  -\sqrt{y} \leq x \leq \sqrt{y}
  \]

### Step 3: Set up the triple integral
The volume \( V \) can be expressed as:
\[
V = \int_{y=0}^{18} \int_{x=-\sqrt{y}}^{\sqrt{y}} \int_{z=0}^{18-y} dz \, dx \, dy
\]

### Step 4: Evaluate the triple integral
Start by evaluating the integral with respect to \( z \):
\[
\int_{0}^{18-y} dz = \left[ z \right]_{0}^{18-y} = 18 - y
\]
Now substitute this into the integral:
\[
V = \int_{y=0}^{18} \int_{x=-\sqrt{y}}^{\sqrt{y}} (18 - y) \, dx \, dy
\]
Next, evaluate the integral with respect to \( x \):
\[
\int_{-\sqrt{y}}^{\sqrt{y}} (18 - y) \, dx = (18 - y) \int_{-\sqrt{y}}^{\sqrt{y}} dx = (18 - y) \left[ x \right]_{-\sqrt{y}}^{\sqrt{y}} = (18 - y) \times (2\sqrt{y})
\]
So the volume integral becomes:
\[
V = \int_{y=0}^{18} 2\sqrt{y}(18 - y) \, dy
\]
### Step 5: Simplify and compute the final integral
Simplify the integrand and integrate with respect to \( y \):
\[
V = 2 \int_{y=0}^{18} \left(18\sqrt{y} - y^{3/2}\right) dy
\]
Evaluate each term separately:
- The integral of \( 18\sqrt{y} \) with respect to \( y \):
  \[
  \int 18\sqrt{y} \, dy = 18 \times \frac{2}{3} y^{3/2} = 12y^{3/2}
  \]
- The integral of \( y^{3/2} \) with respect to \( y \):
  \[
  \int y^{3/2} \, dy = \frac{2}{5} y^{5/2}
  \]

Thus, the integral becomes:
\[
V = 2 \left[ 12y^{3/2} - \frac{2}{5} y^{5/2} \right]_{0}^{18}
\]
Substitute \( y = 18 \) into the expression:
\[
V = 2 \left[ 12(18)^{3/2} - \frac{2}{5} (18)^{5/2} \right]
\]
Simplify and compute the final result:
\[
(18)^{3/2} = 18 \times \sqrt{18} = 18 \times 3\sqrt{2} = 54\sqrt{2}
\]
\[
(18)^{5/2} = 18 \times (18)^{3/2} = 18 \times 54\sqrt{2} = 972\sqrt{2}
\]
\[
V = 2 \left[ 12 \times 54\sqrt{2} - \frac{2}{5} \times 972\sqrt{2} \right]
\]
\[
V = 2 \times \left[ 648\sqrt{2} - \frac{1944\sqrt{2}}{5} \right]
\]
Simplify further:
\[
V = 2 \times \left( \frac{3240\sqrt{2}}{5} - \frac{1944\sqrt{2}}{5} \right) = 2 \times \frac{1296\sqrt{2}}{5} = \frac{2592\sqrt{2}}{5}
\]
Therefore, the volume of the wedge is:
\[
V = \frac{2592\sqrt{2}}{5} \text{ cubic units}
\]

Would you like to ask anything about the procedure or get more details?

---

1. How do you approach setting up triple integrals for volume calculation in other scenarios?
2. Can the order of integration be changed in this problem?
3. How would cylindrical coordinates simplify the evaluation of certain integrals?
4. What are the general steps for finding the bounds in a triple integral problem?
5. How would the problem change if the planes were different?

**Tip:** When dealing with complex integrals, it's helpful to visualize the region of integration first to ensure that the bounds are correctly identified.

Learn how to calculate the volume of a wedge bounded by the parabolic cylinder y = x^2 and the planes z = 18 - y and z = 0 using triple integrals. Detailed step-by-step solution with bounds determination and integral evaluation.

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