# Step 1: Understand the Surface and the Vector Field Let us first analyze the surface \( S \) and the vector field: - The surface \( S \) is the portion of the plane \( x + y + z = 1 \) in the **first octant** (\( x, y, z \geq \)), together with the coordinate planes \( x= \), \( y= \), and \( z= \) that bound it. - This forms a tetrahedron with vertices at \((,,)\), \((1,,)\), \((,1,)\), and \((,,1)\). - The vector field is given by \(\vec{F} = x^2 \mathbf{i} + y^2 \mathbf{j} - xy^3 \mathbf{k}\). - The normal vector \(\mathbf{\hat{n}}\) is **inward-pointing**, which is opposite to the usual outward-pointing normal for the divergence theorem. **Explanation:** Understanding the domain and the orientation of the normal is vital, as applying the divergence theorem requires attention to the normal's direction. Since the problem asks for inward orientation, we must adjust the sign at the end if we use the standard divergence theorem. --- # Step 2: Apply the Divergence Theorem The **divergence theorem** relates the surface integral of a vector field over a closed surface to the triple integral of the divergence over the volume inside the surface: \[ \oiint_{S} \vec{F} \cdot \mathbf{\hat{n}}\,dS = \iiint_{V} (\nabla \cdot \vec{F})\, dV \] But, since the divergence theorem assumes an **outward** normal, and our problem requires the **inward** normal, the value will be the **negative** of the usual divergence theorem result. Now, calculate the divergence: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(-xy^3) = 2x + 2y + = 2(x + y) \] **Explanation:** By computing the divergence, we simplify the surface integral to a volume integral. Noting the need to reverse the sign at the end is crucial due to the inward orientation. --- # Step 3: Set Up and Compute the Triple Integral The region \( V \) is the tetrahedron defined earlier. Its limits are: - \( x \) goes from \( \) to \( 1 \) - \( y \) goes from \( \) to \( 1 - x \) - \( z \) goes from \( \) to \( 1 - x - y \) So, \[ \iiint_{V} 2(x + y)\, dV = 2 \int_{x=}^{1} \int_{y=}^{1-x} \int_{z=}^{1-x-y} (x + y)\, dz\, dy\, dx \] Integrate over \( z \): \[ \int_{z=}^{1-x-y} dz = 1 - x - y \] So the integral becomes: \[ 2 \int_{x=}^{1} \int_{y=}^{1-x} (x + y)(1 - x - y)\, dy\, dx \] Expand the integrand: \[ (x + y)(1 - x - y) = x(1 - x - y) + y(1 - x - y) = x(1 - x) - x y + y(1 - x) - y^2 = x - x^2 - x y + y - x y - y^2 \] \[ = x - x^2 - 2 x y + y - y^2 \] Now, integrate with respect to \( y \): \[ \int_{y=}^{1-x} [x - x^2 - 2 x y + y - y^2]\, dy \] Integrate term by term: - \( \int_{y=}^{1-x} x\, dy = x(1-x) \) - \( \int_{y=}^{1-x} -x^2\, dy = -x^2 (1-x) \) - \( \int_{y=}^{1-x} -2 x y\, dy = -2x \int_{}^{1-x} y\, dy = -2x \cdot \frac{1}{2}(1-x)^2 = -x (1-x)^2 \) - \( \int_{y=}^{1-x} y\, dy = \frac{1}{2}(1-x)^2 \) - \( \int_{y=}^{1-x} -y^2\, dy = -\frac{1}{3}(1-x)^3 \) Sum: \[ x(1-x) - x^2(1-x) - x(1-x)^2 + \frac{1}{2}(1-x)^2 - \frac{1}{3}(1-x)^3 \] Now, integrate over \( x \) from to 1. Let’s combine and write the full integral: \[ 2 \int_{x=}^{1} \left[ x(1-x) - x^2(1-x) - x(1-x)^2 + \frac{1}{2}(1-x)^2 - \frac{1}{3}(1-x)^3 \right] dx \] Let's simplify each term: - \( x(1-x) = x - x^2 \) - \( -x^2(1-x) = -x^2 + x^3 \) - \( -x(1-x)^2 = -x(1 - 2x + x^2) = -x + 2x^2 - x^3 \) - \( \frac{1}{2}(1-x)^2 = \frac{1}{2} (1 - 2x + x^2) = \frac{1}{2} - x + \frac{1}{2} x^2 \) - \( -\frac{1}{3}(1-x)^3 = -\frac{1}{3}(1 - 3x + 3x^2 - x^3) = -\frac{1}{3} + x - x^2 + \frac{1}{3} x^3 \) Add all: \[ \begin{align*} & [x - x^2] + [-x^2 + x^3] + [-x + 2x^2 - x^3] + [\frac{1}{2} - x + \frac{1}{2} x^2] + [-\frac{1}{3} + x - x^2 + \frac{1}{3} x^3] \\ = & (x - x^2 - x^2 + x^3 - x + 2x^2 - x^3 + \frac{1}{2} - x + \frac{1}{2} x^2 - \frac{1}{3} + x - x^2 + \frac{1}{3} x^3) \\ \end{align*} \] Now, combine like terms: - \( x \): \( x - x + (-x) + x = \) - \( x^2 \): \( -x^2 - x^2 + 2x^2 + \frac{1}{2} x^2 - x^2 = (-x^2 - x^2 - x^2) + 2x^2 + \frac{1}{2} x^2 = (-3x^2) + 2x^2 + \frac{1}{2} x^2 = (-x^2) + \frac{1}{2} x^2 = -\frac{1}{2} x^2 \) - \( x^3 \): \( x^3 - x^3 + \frac{1}{3} x^3 = + \frac{1}{3} x^3 = \frac{1}{3} x^3 \) - constants: \( \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \) So, the integrand is: \[ -\frac{1}{2} x^2 + \frac{1}{3} x^3 + \frac{1}{6} \] Now, integrate from \( x= \) to \( x=1 \): \[ 2 \int_{}^{1} \left( -\frac{1}{2} x^2 + \frac{1}{3} x^3 + \frac{1}{6} \right) dx \] Compute each term: - \( \int_{}^{1} x^2 dx = \left. \frac{1}{3} x^3 \right|_^1 = \frac{1}{3} \) - \( \int_{}^{1} x^3 dx = \left. \frac{1}{4} x^4 \right|_^1 = \frac{1}{4} \) - \( \int_{}^{1} dx = 1 \) Plug in: \[ 2 \left( -\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{6} \cdot 1 \right ) = 2 \left( -\frac{1}{6} + \frac{1}{12} + \frac{1}{6} \right ) \] Now sum inside the parentheses: - \( -\frac{1}{6} + \frac{1}{6} = \) - So, \( + \frac{1}{12} = \frac{1}{12} \) Multiply by 2: \[ 2 \times \frac{1}{12} = \frac{1}{6} \] But recall, the original surface integral with **inward normal** is the **negative** of this result. --- # Final Answer \[ \boxed{-\frac{1}{6}} \] **Explanation:** By carefully applying the divergence theorem and accounting for the inward normal, we find the value of the given surface integral is \( -\frac{1}{6} \). Each step, from understanding the region and vector field, through setting up the triple integral, to detailed calculation, ensures the result is precise and justified.