We are tasked with finding the volume of a wedge bounded by the parabolic cylinder $y = x^2$, the plane $z = 18 - y$, and the plane $z = 0$. To do this, we'll set up a triple integral in the order $dz\,dy\,dx$.

Step 1: Determine the limits of integration

1. For $z$:
   The wedge is bounded by the planes $z = 0$ and $z = 18 - y$, so the limits for $z$ are: $0 \leq z \leq 18 - y$

2. For $y$:
   Since $y = x^2$, we need to determine the range for $y$. The boundary of $y$ is from 0 to 18 because the plane $z = 18 - y$ implies that $y \leq 18$. Hence, the limits for $y$ are: $0 \leq y \leq 18$

3. For $x$:
   Since $y = x^2$, for each $y$, $x$ ranges from $-\sqrt{y}$ to $\sqrt{y}$. So the limits for $x$ are: $-\sqrt{y} \leq x \leq \sqrt{y}$

Step 2: Set up the triple integral

The volume $V$ is given by the triple integral:

$V = \int_{y=0}^{18} \int_{x=-\sqrt{y}}^{\sqrt{y}} \int_{z=0}^{18-y} dz \, dx \, dy$

Step 3: Evaluate the integrals

1. Integrating with respect to $z$: $\int_{0}^{18-y} dz = (18 - y) - 0 = 18 - y$

   So the integral becomes: $V = \int_{y=0}^{18} \int_{x=-\sqrt{y}}^{\sqrt{y}} (18 - y) \, dx \, dy$

2. Integrating with respect to $x$: Since the integrand $18 - y$ does not depend on $x$, we can integrate $x$ directly: $\int_{-\sqrt{y}}^{\sqrt{y}} dx = 2\sqrt{y}$

   Now the integral becomes: $V = \int_{y=0}^{18} (18 - y) \cdot 2\sqrt{y} \, dy$ $V = 2 \int_{y=0}^{18} (18 - y) \sqrt{y} \, dy$

3. Integrating with respect to $y$:

We now need to integrate $(18 - y) \sqrt{y}$. We can break this into two terms: $V = 2 \left( 18 \int_{y=0}^{18} \sqrt{y} \, dy - \int_{y=0}^{18} y^{3/2} \, dy \right)$

• For the first term: $\int \sqrt{y} \, dy = \frac{2}{3} y^{3/2}$ So: $18 \int_{0}^{18} \sqrt{y} \, dy = 18 \cdot \frac{2}{3} \cdot 18^{3/2} = 12 \cdot 18^{3/2}$

• For the second term: $\int y^{3/2} \, dy = \frac{2}{5} y^{5/2}$ So: $\int_{0}^{18} y^{3/2} \, dy = \frac{2}{5} \cdot 18^{5/2}$

Now substituting back, we get: $V = 2 \left( 12 \cdot 18^{3/2} - \frac{2}{5} \cdot 18^{5/2} \right)$

Step 4: Simplify and find the volume

• $18^{3/2} = 18 \cdot \sqrt{18} = 18 \cdot 4.2426 \approx 76.3668$
• $18^{5/2} = 18^2 \cdot \sqrt{18} = 324 \cdot 4.2426 \approx 1374.6$

So: $V = 2 \left( 12 \cdot 76.3668 - \frac{2}{5} \cdot 1374.6 \right)$ $V \approx 2 \left( 916.4016 - 549.84 \right)$ $V \approx 2 \times 366.5616 = 733.1232$

Thus, the volume of the wedge is approximately $\boxed{733.1232}$ cubic units.

Would you like more details, or have any questions?

Related Questions:

1. How would the volume change if the plane equation was $z = 20 - y$?
2. Can you set up the same integral using cylindrical coordinates?
3. What would be the volume if the cylinder was defined as $y = 2x^2$?
4. How does the symmetry in $x$ affect the limits and computation of this volume?
5. How do you generalize the setup of a triple integral for finding volumes in different regions?

Tip:

When setting up triple integrals, carefully analyze the boundaries of each variable to ensure the integration limits are correct.