To determine the volume of the solid bounded by the given surfaces in the first octant, let's analyze the problem step by step:

Given surfaces:

1. $y + z = 3$: A plane.
2. $y = x^2$: A parabolic surface.
3. $xy$-plane: This is the plane $z = 0$.
4. $yz$-plane: This is the plane $x = 0$.

The first octant restricts $x, y, z \geq 0$.

Plan:

We calculate the volume by setting up a triple integral: $V = \int \int \int_{R} dz \, dy \, dx$ where $R$ represents the region bounded by the above constraints. We'll compute $z$ first (based on $y + z = 3$), followed by $y$ (bounded by $y = x^2$), and then $x$.

Step 1: Limits of integration

1. $z$-limits: From $z = 0$ (the $xy$-plane) to $z = 3 - y$ (from $y + z = 3$). $0 \leq z \leq 3 - y$

2. $y$-limits: $y$ is bounded by $y = x^2$ and the intersection of $y + z = 3$ with the $xy$-plane ($z = 0$, so $y = 3$). $x^2 \leq y \leq 3$

3. $x$-limits: $x$ is bounded by $x = 0$ (from the $yz$-plane) and $x = \sqrt{y}$ (from $y = x^2$). $0 \leq x \leq \sqrt{y}$

Step 2: Write the triple integral

The volume $V$ is: $V = \int_{x=0}^{\sqrt{y}} \int_{y=x^2}^{3} \int_{z=0}^{3-y} dz \, dy \, dx$

Step 3: Evaluate the integral

Step 3.1: Integrate with respect to $z$

$\int_{z=0}^{3-y} dz = [z]_{z=0}^{z=3-y} = (3-y)$ The integral becomes: $V = \int_{x=0}^{\sqrt{y}} \int_{y=x^2}^{3} (3-y) \, dy \, dx$

Step 3.2: Integrate with respect to $y$

$\int_{y=x^2}^{3} (3-y) \, dy = \int_{y=x^2}^{3} 3 \, dy - \int_{y=x^2}^{3} y \, dy$ Evaluate each term:

1. $\int_{y=x^2}^{3} 3 \, dy = 3[y]_{y=x^2}^{y=3} = 3(3 - x^2)$
2. $\int_{y=x^2}^{3} y \, dy = \left[\frac{y^2}{2}\right]_{y=x^2}^{y=3} = \frac{3^2}{2} - \frac{(x^2)^2}{2} = \frac{9}{2} - \frac{x^4}{2}$

Combine these: $\int_{y=x^2}^{3} (3-y) \, dy = 3(3 - x^2) - \left(\frac{9}{2} - \frac{x^4}{2}\right) = \frac{9}{2} - 3x^2 + \frac{x^4}{2}$

The integral becomes: $V = \int_{x=0}^{\sqrt{3}} \left(\frac{9}{2} - 3x^2 + \frac{x^4}{2}\right) dx$

Step 3.3: Integrate with respect to $x$

Split into separate terms: $\int_{x=0}^{\sqrt{3}} \frac{9}{2} dx = \frac{9}{2}x \Big|_{0}^{\sqrt{3}} = \frac{9}{2} \sqrt{3}$ $\int_{x=0}^{\sqrt{3}} -3x^2 dx = -3 \cdot \frac{x^3}{3} \Big|_{0}^{\sqrt{3}} = -\sqrt{3}^3 = -3\sqrt{3}$ $\int_{x=0}^{\sqrt{3}} \frac{x^4}{2} dx = \frac{1}{2} \cdot \frac{x^5}{5} \Big|_{0}^{\sqrt{3}} = \frac{1}{2} \cdot \frac{(\sqrt{3})^5}{5} = \frac{3^{5/2}}{10}$

Combine these: $V = \frac{9}{2}\sqrt{3} - 3\sqrt{3} + \frac{3^{5/2}}{10}$

Simplify: $V = \frac{9\sqrt{3}}{2} - \frac{6\sqrt{3}}{2} + \frac{3^{5/2}}{10}$ $V = \frac{3\sqrt{3}}{2} + \frac{3^{5/2}}{10}$

Final Answer:

$V = \frac{3\sqrt{3}}{2} + \frac{3^{5/2}}{10}$

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Would you like a more detailed breakdown of any step?

Related Questions:

1. How do you determine the bounds of integration for a solid in the first octant?
2. What is the geometric interpretation of $y = x^2$ and $y + z = 3$?
3. Can you set up a similar problem with different bounding surfaces?
4. How does symmetry simplify triple integrals?
5. How does the order of integration affect the solution?

Tip:

Always visualize the bounding surfaces and intersection curves for clarity in setting up the integral.