Let's solve the given flux integral step-by-step using the **Divergence Theorem**. --- ### **Step 1: Write the Vector Field and the Region** Given: \[ \vec{F}(x, y, z) = xy\,\vec{i} + (y^2 + e^{xz^2})\,\vec{j} + \sin(xy)\,\vec{k} \] The region \( D \) is bounded by: - \( x = 1 - x^2 \) (which means \( x \) goes from \(-1\) to \(1\)) - \( y = 0 \) - \( z = 0 \) - \( y + z = 2 \) So, the boundaries in terms of variables: - \( x \) from \(-1\) to \(1\) - \( y \) from \(0\) to \(1 - x^2\) (from the parabolic cylinder) - \( z \) from \(0\) to \(2-y\) (from the planes \( z=0 \) and \( y+z=2 \)) --- ### **Step 2: State the Divergence Theorem** The Divergence Theorem states: \[ \iint_{S} \vec{F} \cdot \vec{n} \, dS = \iiint_{D} (\nabla \cdot \vec{F}) \, dV \] --- ### **Step 3: Compute the Divergence** \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(y^2 + e^{xz^2}) + \frac{\partial}{\partial z}(\sin(xy)) \] Calculate each term: - \(\frac{\partial}{\partial x}(xy) = y\) - \(\frac{\partial}{\partial y}(y^2 + e^{xz^2}) = 2y\) - \(\frac{\partial}{\partial z}(\sin(xy)) = 0\) (since \(\sin(xy)\) does not depend on \(z\)) So, \[ \nabla \cdot \vec{F} = y + 2y = 3y \] --- ### **Step 4: Set Up the Triple Integral** \[ \iiint_{D} 3y \, dV \] Bounds (as described above): \[ \int_{x=-1}^{1} \int_{y=0}^{1-x^2} \int_{z=0}^{2-y} 3y \, dz\,dy\,dx \] Integrate with respect to \(z\) first: \[ \int_{z=0}^{2-y} dz = (2-y) - 0 = (2-y) \] So the integral becomes: \[ \int_{x=-1}^{1} \int_{y=0}^{1-x^2} 3y(2-y)\, dy\, dx \] \[ = \int_{x=-1}^{1} \int_{y=0}^{1-x^2} (6y - 3y^2)\, dy\, dx \] --- ### **Step 5: Integrate with Respect to \(y\)** Integrate \(6y - 3y^2\) with respect to \(y\): \[ \int (6y - 3y^2)\, dy = 3y^2 - y^3 \] Evaluate from \(y=0\) to \(y=1-x^2\): \[ = 3(1 - x^2)^2 - (1 - x^2)^3 \] --- ### **Step 6: Integrate with Respect to \(x\)** Now set up the last integral: \[ \int_{x=-1}^{1} \left[ 3(1 - x^2)^2 - (1 - x^2)^3 \right] dx \] Let's expand: - \((1 - x^2)^2 = 1 - 2x^2 + x^4\) - \((1 - x^2)^3 = 1 - 3x^2 + 3x^4 - x^6\) So, \[ 3(1 - x^2)^2 - (1 - x^2)^3 = 3(1 - 2x^2 + x^4) - [1 - 3x^2 + 3x^4 - x^6] \] \[ = (3 - 6x^2 + 3x^4) - (1 - 3x^2 + 3x^4 - x^6) \] \[ = 3 - 6x^2 + 3x^4 - 1 + 3x^2 - 3x^4 + x^6 \] \[ = (3 - 1) + (-6x^2 + 3x^2) + (3x^4 - 3x^4) + x^6 \] \[ = 2 - 3x^2 + x^6 \] So, our integral is: \[ \int_{x=-1}^{1} (2 - 3x^2 + x^6) dx \] --- ### **Step 7: Evaluate the Final Integral** Each term is even, so double the integral from 0 to 1: \[ = 2 \int_{0}^{1} (2 - 3x^2 + x^6) dx \] \[ = 2 \left[ 2x - x^3 + \frac{1}{7}x^7 \right]_{0}^{1} \] \[ = 2 \left[ 2(1) - (1)^3 + \frac{1}{7}(1)^7 - (0 - 0 + 0) \right] \] \[ = 2 \left[ 2 - 1 + \frac{1}{7} \right] \] \[ = 2 \left[ 1 + \frac{1}{7} \right] \] \[ = 2 \cdot \frac{8}{7} = \frac{16}{7} \] --- ### **Final Answer** \[ \boxed{\frac{16}{7}} \] That is the value of the surface integral \(\iint_S \vec{F} \cdot \vec{n} \, dS\) over the given region.