Question

Let C be the curve of intersection of the cylinder \( x^2 + y^2 = 4 \) and the plane \( z - 2 = 0 \). Suppose C is oriented in the counterclockwise direction around the z-axis, when viewed from above. If \[ \int_C ( \sin x + e^x ) dx + 4x dy + e^z \cos^2 z dz = \alpha\pi, \] then the value of \( \alpha \) equals ..................

Hint:

Whenever you see a line integral over a simple closed curve (like a circle or ellipse), always check if Stokes' Theorem can simplify the problem. If the curl of the vector field is simple (e.g., constant or zero), the surface integral will be much easier to compute than parameterizing the curve and doing the line integral directly.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the evaluation of a line integral over a closed curve \( C \) in 3D space. The curve \( C \) is a circle. The presence of a closed curve and a vector field suggests that Stokes' Theorem might be a simpler approach than direct parameterization.
Step 2: Key Formula or Approach:
Stokes' Theorem states that for a vector field \( \mathbf{F} \) and an oriented surface \( S \) with boundary curve \( C = \partial S \), the line integral of \( \mathbf{F} \) around \( C \) is equal to the surface integral of the curl of \( \mathbf{F} \) over \( S \): \[ \oint_C \mathbf{F} . d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) . \mathbf{n} \, dS \] The vector field is \( \mathbf{F}(x, y, z) = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \), where:
\( P = \sin x + e^x \)
\( Q = 4x \)
\( R = e^z \cos^2 z \)
The curl is \( \nabla \times \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathbf{i} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right)\mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k} \).
Step 3: Detailed Explanation or Calculation:
1. Identify the Curve and Surface:
The curve \( C \) is the intersection of the cylinder \( x^2 + y^2 = 4 \) and the plane \( z = 2 \). This is a circle of radius 2, centered at (0,0,2) in the plane \( z=2 \). The surface \( S \) bounded by \( C \) is the disk \( x^2 + y^2 \le 4 \) in the plane \( z=2 \).
The orientation of \( C \) is counterclockwise when viewed from above. By the right-hand rule, the normal vector \( \mathbf{n} \) to the surface \( S \) must point upwards. The plane is \( z=2 \), so the upward normal is \( \mathbf{n} = \mathbf{k} = (0, 0, 1) \).
2. Calculate the Curl of F:
We compute the partial derivatives: \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(\sin x + e^x) = 0 \] \[ \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(\sin x + e^x) = 0 \] \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x) = 4 \] \[ \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(4x) = 0 \] \[ \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(e^z \cos^2 z) = 0 \] \[ \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(e^z \cos^2 z) = 0 \] Now, assemble the curl vector: \[ \nabla \times \mathbf{F} = (0 - 0)\mathbf{i} + (0 - 0)\mathbf{j} + (4 - 0)\mathbf{k} = \langle 0, 0, 4 \rangle \] 3. Evaluate the Surface Integral:
The integral becomes: \[ \iint_S (\nabla \times \mathbf{F}) . \mathbf{n} \, dS = \iint_S \langle 0, 0, 4 \rangle . \langle 0, 0, 1 \rangle \, dS \] \[ = \iint_S 4 \, dS \] This is simply 4 times the area of the surface \( S \). The surface \( S \) is a disk with radius \( r=2 \). The area of \( S \) is \( \pi r^2 = \pi (2)^2 = 4\pi \). So, the value of the integral is: \[ 4 \times (\text{Area of } S) = 4 \times 4\pi = 16\pi \] 4. Find the value of \( \alpha \):
We are given that the integral equals \( \alpha\pi \). \[ 16\pi = \alpha\pi \] Therefore, \( \alpha = 16 \).
Step 4: Final Answer:
The value of \( \alpha \) is 16.
Step 5: Why This is Correct:
The calculation using Stokes' theorem is straightforward. The curl of the vector field is a constant vector pointing in the z-direction. The surface of integration is a flat disk in a plane parallel to the xy-plane, making its normal vector also a constant in the z-direction. This simplifies the surface integral to a constant multiplied by the area of the disk, leading to the result \( 16\pi \).