Evaluate the surface integral || F. ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = -xi – yj + z³k, S is the part of the cone z = V x2 between the planes z = 1 and z = 2 with downward orientation

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Please solve the screenshot. The answer is none of 109pi/15, -109pi/15, 209pi, or -209pi/15. Thanks. 

**Evaluate the Surface Integral**

Evaluate the surface integral 

\[
\iint_S \mathbf{F} \cdot d\mathbf{S}
\]

for the given vector field **F** and the oriented surface **S**. In other words, find the flux of **F** across **S**. For closed surfaces, use the positive (outward) orientation.

**Vector Field:**
\[
\mathbf{F}(x, y, z) = -x\mathbf{i} - y\mathbf{j} + z^3\mathbf{k}
\]

**Surface:**

**S** is the part of the cone 

\[
z = \sqrt{x^2 + y^2}
\]

between the planes \( z = 1 \) and \( z = 2 \) with downward orientation.
Transcribed Image Text:**Evaluate the Surface Integral** Evaluate the surface integral \[ \iint_S \mathbf{F} \cdot d\mathbf{S} \] for the given vector field **F** and the oriented surface **S**. In other words, find the flux of **F** across **S**. For closed surfaces, use the positive (outward) orientation. **Vector Field:** \[ \mathbf{F}(x, y, z) = -x\mathbf{i} - y\mathbf{j} + z^3\mathbf{k} \] **Surface:** **S** is the part of the cone \[ z = \sqrt{x^2 + y^2} \] between the planes \( z = 1 \) and \( z = 2 \) with downward orientation.
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