Evaluate the surface integral $\iint_S yz \, dS$ where $S$ is the part of the plane $x + y + z = 4$ that lies in the first octant.

Neetesh Kumar | November 10, 2024

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This is the solution to Math 1D
Assignment: 16.7 Question Number 3
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Step-by-step solution:

To evaluate the surface integral $\iint_S yz \, dS$, we need to parameterize the surface $S$ and express $dS$ in terms of $dx$ and $dy$.

Step 1: Find the Equation of the Plane

The equation of the plane is given by: $x + y + z = 4$ Solving for $z$ in terms of $x$ and $y$, we get: $z = 4 - x - y$

Step 2: Set Up the Surface Element $dS$

The surface element $dS$ for a plane $z = f(x, y)$ is given by: $dS = \sqrt{\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 + 1} \, dx \, dy$

Calculate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:

1. $\frac{\partial z}{\partial x} = -1$
2. $\frac{\partial z}{\partial y} = -1$

Thus, $dS = \sqrt{(-1)^2 + (-1)^2 + 1} \, dx \, dy = \sqrt{3} \, dx \, dy$

Step 3: Set Up the Integral

Now we can express the integral as: $\iint_S yz \, dS = \iint_D y(4 - x - y) \sqrt{3} \, dx \, dy$

where $D$ is the projection of $S$ onto the $xy$-plane, which is a triangular region with vertices at $(4, 0)$, $(0, 4)$, and $(0, 0)$.

Step 4: Set Up the Limits of Integration

For the triangular region $D$ in the $xy$-plane:

• $x$ ranges from $0$ to $4$.
• For a given $x$, $y$ ranges from $0$ to $4 - x$.

Thus, the integral becomes: $\iint_S yz \, dS = \sqrt{3} \int_0^4 \int_0^{4 - x} y(4 - x - y) \, dy \, dx$

Step 5: Evaluate the Integral

1. Integrate with respect to $y$: $\int_0^{4 - x} y(4 - x - y) \, dy = \int_0^{4 - x} (4y - xy - y^2) \, dy$

   Break it down term by term:

   • $\int_0^{4 - x} 4y \, dy = 4 \cdot \frac{(4 - x)^2}{2} = 2(4 - x)^2$
   • $\int_0^{4 - x} xy \, dy = x \cdot \frac{(4 - x)^2}{2} = \frac{x(4 - x)^2}{2}$
   • $\int_0^{4 - x} y^2 \, dy = \frac{(4 - x)^3}{3}$

   Putting these together: $\int_0^{4 - x} y(4 - x - y) \, dy = 2(4 - x)^2 - \frac{x(4 - x)^2}{2} - \frac{(4 - x)^3}{3}$

2. Integrate with respect to $x$: Substitute back into the outer integral and evaluate from $0$ to $4$:

   $\iint_S yz \, dS = \sqrt{3} \int_0^4 \left( 2(4 - x)^2 - \frac{x(4 - x)^2}{2} - \frac{(4 - x)^3}{3} \right) dx = \dfrac{32}{\sqrt{3}}$

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Final Answer:

$\iint_S yz \, dS = \boxed{\dfrac{32}{\sqrt{3}}}$

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