To find the volume of the solid region in the first octant bounded by the plane $2x + 5y + 10z = 1$ and the coordinate planes using triple integrals, we proceed as follows:

Step 1: Solve for bounds

1. Plane equation intersection with axes:

   • When $y = 0$ and $z = 0$: $2x = 1 \implies x = \frac{1}{2}$.
   • When $x = 0$ and $z = 0$: $5y = 1 \implies y = \frac{1}{5}$.
   • When $x = 0$ and $y = 0$: $10z = 1 \implies z = \frac{1}{10}$.

   Thus, the triangular base is formed by the intersection of the plane with the coordinate planes.

2. Plane equation in terms of $z$: Solving for $z$: $z = \frac{1 - 2x - 5y}{10}.$

3. Bounds:

   • $x$ ranges from $0$ to $\frac{1}{2}$.
   • For a fixed $x$, $y$ ranges from $0$ to $\frac{1 - 2x}{5}$.
   • For fixed $x$ and $y$, $z$ ranges from $0$ to $\frac{1 - 2x - 5y}{10}$.

Step 2: Triple Integral Setup

The volume $V$ is given by: $V = \int_0^{\frac{1}{2}} \int_0^{\frac{1 - 2x}{5}} \int_0^{\frac{1 - 2x - 5y}{10}} 1 \, dz \, dy \, dx.$

Step 3: Compute the Integral

1. Inner integral with respect to $z$: $\int_0^{\frac{1 - 2x - 5y}{10}} 1 \, dz = \left[ z \right]_0^{\frac{1 - 2x - 5y}{10}} = \frac{1 - 2x - 5y}{10}.$

   The integral reduces to: $V = \int_0^{\frac{1}{2}} \int_0^{\frac{1 - 2x}{5}} \frac{1 - 2x - 5y}{10} \, dy \, dx.$

2. Second integral with respect to $y$: Expand the numerator: $\int_0^{\frac{1 - 2x}{5}} \frac{1 - 2x - 5y}{10} \, dy = \frac{1}{10} \int_0^{\frac{1 - 2x}{5}} (1 - 2x) \, dy - \frac{1}{10} \int_0^{\frac{1 - 2x}{5}} 5y \, dy.$

   Evaluate each term:

   • For $\int_0^{\frac{1 - 2x}{5}} (1 - 2x) \, dy$: $(1 - 2x) \int_0^{\frac{1 - 2x}{5}} 1 \, dy = (1 - 2x) \cdot \frac{1 - 2x}{5} = \frac{(1 - 2x)^2}{5}.$

   • For $\int_0^{\frac{1 - 2x}{5}} 5y \, dy$: $\int_0^{\frac{1 - 2x}{5}} 5y \, dy = 5 \cdot \frac{y^2}{2} \Big|_0^{\frac{1 - 2x}{5}} = 5 \cdot \frac{\left(\frac{1 - 2x}{5}\right)^2}{2} = \frac{(1 - 2x)^2}{10}.$

   Combine the results: $\frac{1}{10} \int_0^{\frac{1 - 2x}{5}} (1 - 2x - 5y) \, dy = \frac{1}{10} \left( \frac{(1 - 2x)^2}{5} - \frac{(1 - 2x)^2}{10} \right) = \frac{1}{10} \cdot \frac{(1 - 2x)^2}{10} = \frac{(1 - 2x)^2}{100}.$

   The integral reduces to: $V = \int_0^{\frac{1}{2}} \frac{(1 - 2x)^2}{100} \, dx.$

3. Final integral with respect to $x$: Expand $(1 - 2x)^2$: $(1 - 2x)^2 = 1 - 4x + 4x^2.$

   Substitute and integrate term by term: $\int_0^{\frac{1}{2}} \frac{1 - 4x + 4x^2}{100} \, dx = \frac{1}{100} \left[ x - 2x^2 + \frac{4x^3}{3} \right]_0^{\frac{1}{2}}.$

   Evaluate at $x = \frac{1}{2}$: $\frac{1}{100} \left( \frac{1}{2} - 2 \cdot \frac{1}{4} + \frac{4}{3} \cdot \frac{1}{8} \right) = \frac{1}{100} \left( \frac{1}{2} - \frac{1}{2} + \frac{1}{6} \right) = \frac{1}{100} \cdot \frac{1}{6} = \frac{1}{600}.$

Final Answer:

The volume of the solid is: $V = \frac{1}{600}.$

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Would you like a detailed explanation for any step?

Related Questions:

1. How do you determine integration bounds for solids in triple integrals?
2. What is the geometric interpretation of this problem?
3. Can you compute the volume using other coordinate systems (like cylindrical)?
4. How can you verify the result using computational tools?
5. What are other applications of triple integrals in physics and engineering?

Tip:

Always visualize the solid by sketching or using software to confirm the bounds before setting up triple integrals.