Evaluating Surface Integrals of Vector Fields Over a Hemisphere: A Comprehensive Guide

In this article, we will provide a step-by-step guide to evaluate the surface integral of a vector field over a hemisphere. The specific vector field we will consider is (mathbf{F} ymathbf{i} - xmathbf{j} 4zmathbf{k}), and the hemisphere is defined by the equation (x^2 y^2 z^2 4, z geq 0). The orientation of the hemisphere is downward.

Introduction to Surface Integrals and Vector Fields

A surface integral of a vector field over a surface measures the flux of the vector field across that surface. This concept is crucial in vector calculus and has numerous applications in physics and engineering.

Parameterizing the Surface of the Hemisphere

To evaluate the surface integral, we first need to parameterize the surface. For a hemisphere, we can use spherical coordinates. Specifically, we use the parameterization:

x rsinthetacosphi, y rsinthetasinphi, and z rcostheta, where r 2, 0 leq theta leq frac{pi}{2}, and 0 leq phi leq 2pi.

Calculating the Normal Vector

The next step is to find the outward unit normal vector to the surface. The outward unit normal vector is given by the cross product of the partial derivatives of the parameterization with respect to (theta) and (phi).

The partial derivatives are:

(frac{partial mathbf{r}}{partial theta} 2costhetacosphimathbf{i} 2costhetasinphimathbf{j} - 2sinthetamathbf{k}) (frac{partial mathbf{r}}{partial phi} -2sinthetasinphimathbf{i} 2sinthetacosphimathbf{j})

The cross product of these vectors yields the normal vector:

(mathbf{N} -4costhetacosphimathbf{i} - 4costhetasinphimathbf{j} - 4sinthetamathbf{k})

Evaluating (mathbf{F} cdot dmathbf{S})

To proceed, we need to evaluate the vector field (mathbf{F}) and the differential element of surface area (dmathbf{S}).

The vector field in terms of the parameters is:

(mathbf{F}(theta, phi)  ymathbf{i} - xmathbf{j}   4zmathbf{k}  2sinthetasinphicosphimathbf{i} - 2sinthetacosphisinphimathbf{j}   8costhetamathbf{k})

The differential element of surface area is:

(dS |frac{partial mathbf{r}}{partial theta} times frac{partial mathbf{r}}{partial phi}|dtheta dphi 8sintheta dtheta dphi)

Setting Up and Solving the Integral

The surface integral (iint_S mathbf{F} cdot dmathbf{S}) can be set up as follows:

(iint_S mathbf{F} cdot dmathbf{S} int_0^{2pi} int_0^{pi/2} (2sinthetasinphicosphi(-4costhetacosphi) - 2sinthetacosphisinphi(-4costhetasinphi) 8costheta(-4sintheta)) 8sintheta dtheta dphi)

Simplifying the integrand, we get:

(iint_S mathbf{F} cdot dmathbf{S} -16int_0^{2pi} int_0^{pi/2} (sin^2thetacos^2phi sin^2thetacos^2phi - 2sin^2thetacosphisinphi) 8sintheta dtheta dphi)

Since (sin^2thetacos^2phi sin^2thetacos^2phi - 2sin^2thetacosphisinphi 0), the integral simplifies to:

(iint_S mathbf{F} cdot dmathbf{S} 0)

Conclusion

Therefore, the surface integral of the vector field (mathbf{F} ymathbf{i} - xmathbf{j} 4zmathbf{k}) over the hemisphere (x^2 y^2 z^2 4, z geq 0) with downward orientation is (0).

Keywords: surface integral, vector field, hemisphere