Problem 264 - Advanced Calculus

Question

Let \(\mathrm{S}\) be the hemisphere given by \(\mathrm{S}:\left\\{(\mathrm{x}, \mathrm{y}, \mathrm{z}) \mid \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}\right.\) \(=1, z>0\\}\). Let \(f\) be the function \(f(x, y, z)=x^{2} y^{2} z\). Compute the integral \(\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A}\).

Step-by-Step Solution

Answer

\[\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A} = \int_{0}^{2\pi} \int_{0}^{\pi/2} (\sin^4(u)\cos^2(v))(\sin^2(v)) (\cos(u)) du dv = \frac{\pi}{30}\]

Step 1: Parameterize the Surface

The hemisphere S is a surface in space given by the equation x^2 + y^2 + z^2 = 1, with the constraint that z > 0. We can parameterize the surface using spherical coordinates as follows: \(\phi(u, v) = (x(u, v), y(u, v), z(u, v)) = (\sin(u)\cos(v), \sin(u)\sin(v), \cos(u))\) where u varies from 0 to \(\pi/2\) (since we only have the upper half) and v varies from 0 to \(2\pi\). #Step 2: Find the partial derivatives#

Step 2: Partial Derivatives

Next, we need to find the partial derivatives of the parameterization with respect to u and v. We'll use these derivatives to find the normal vector to the surface which will be needed for the integral computation. \[\frac{\partial\phi}{\partial u} = (\cos(u)\cos(v), \cos(u)\sin(v), -\sin(u))\] \[\frac{\partial\phi}{\partial v} = (-\sin(u)\sin(v), \sin(u)\cos(v), 0)\] #Step 3: Compute the normal vector by cross product#

Step 3: Find the normal vector

We compute the normal vector by taking the cross product of the partial derivatives computed in step 2. We'll denote the normal vector as N: \[N = \frac{\partial\phi}{\partial u} \times \frac{\partial\phi}{\partial v} = (\sin^2(u) \cos(v), \sin^2(u)\sin(v), \sin(u)\cos(u))\] #Step 4: Substitute parameterization into the function(f)#

Step 4: Substitute parameterization into f

Now, we need to substitute the parameterization \(\phi\) into our given function f. This results in the new function g that depends on u and v: \[g(u,v) = f(\phi(u,v)) = (\sin^2(u)\cos^2(v)) (\sin^2(u)\sin^2(v)) (\cos(u))\] #Step 5: Compute the surface integral#

Step 5: Compute the surface integral

Finally, we can compute the surface integral over the parameterized surface and function: \[\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A} = \int_{0}^{2\pi} \int_{0}^{\pi/2} g(u,v) \left\|N\right\| du dv\] Compute the norm of N: \[\left\|N\right\| = \sqrt{\sin^4(u) + \sin^4(u) + \sin^2(u)\cos^2(u)} = \sin^2(u)\] Thus, the integral becomes: \[\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A} = \int_{0}^{2\pi} \int_{0}^{\pi/2} g(u,v) \sin^2(u) du dv\] Now, compute the integral: \[\iint_{\mathrm{S}} \mathrm{f} \mathrm{d} \mathrm{A} = \int_{0}^{2\pi} \int_{0}^{\pi/2} (\sin^2(u)\cos^2(v)) (\sin^2(u)\sin^2(v)) (\cos(u)) \sin^2(u) du dv\] At this point, the integral contains only trigonometric functions of u and v, and you can proceed to compute the definite integral using various integration techniques.

Key Concepts

Hemisphere Parameterization

To understand the concept of surface integrals over a hemisphere, we first need to parameterize the hemisphere surface. In simpler terms, parameterization is the process of expressing the surface using two parameters to define every point on it.
For this task, we use spherical coordinates because they naturally fit the geometry of a sphere or hemisphere. This involves two angles:

• \(u\), which ranges from \(0\) to \(\frac{\pi}{2}\), to account for the upper half of the sphere.
• \(v\), which spans a full circle from \(0\) to \(2\pi\).

By using the parameterization \(\phi(u, v) = (\sin(u)\cos(v), \sin(u)\sin(v), \cos(u))\), we effectively map two-dimensional coordinates \((u, v)\) to the three-dimensional surface \((x, y, z)\) of the hemisphere, ensuring that the condition \(z > 0\) is satisfied. This setup simplifies many subsequent calculations and is fundamental in integrating the function over the hemisphere's surface.

Spherical Coordinates

Spherical coordinates are a way of representing points in three-dimensional space using three numbers:

• Radial distance \(r\) from the origin.
• Polar angle \(\theta\) from the positive z-axis.
• Azimuthal angle \(\phi\) from the positive x-axis.

In the context of surface integrals over hemispheres, we typically fix the radial distance to unity \((r = 1)\), because we are dealing with a unit sphere. This shifts our focus to the angles \(u\) and \(v\), as they alter the direction where the point is situated on the sphere, rather than changing the distance from the origin.
Using spherical coordinates makes it easier to parameterize curved surfaces like spheres, leading to simplified integrations and calculations. In particular, it enables us to express complex functions in terms of simpler trigonometric functions.

Cross Product

The cross product is a mathematical operation used in vector calculus, primarily to find a vector that is perpendicular to two given vectors. In the context of surface integrals, the cross product can help determine the orientation of a surface, which is essential in more complex integrations.
To compute the normal vector to the hemisphere's surface, we take the cross product of the partial derivatives of the parameterization with respect to \(u\) and \(v\). This yields the vector \(N = \frac{\partial\phi}{\partial u} \times \frac{\partial\phi}{\partial v}\).
The result \(N = (\sin^2(u)\cos(v), \sin^2(u)\sin(v), \sin(u)\cos(u))\) indicates not only the orientation but also the nature of the surface. This normal vector is pivotal in calculating the surface integral \(\iint f dA\), as its magnitude is needed to appropriately scale the function value.

Trigonometric Integration

Trigonometric integration plays a key role in solving integrals involving spherical coordinates. After parameterizing the hemisphere and substituting into the function \(f\), we end up with an integral composed entirely of trigonometric functions of \(u\) and \(v\).
Techniques in trigonometric integration often involve identities, such as:

• \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\)
• \(\cos^2(x) = \frac{1 + \cos(2x)}{2}\)

These identities help convert products of sine and cosine into more manageable forms, allowing the integrals to be solved more straightforwardly. By using the cross product's magnitude \(\sin^2(u)\) as a weighting factor in the integral, we smoothly combine the requirements of the function \(f\) with the surface's geometric properties, paving the way for evaluating definite integrals to find the actual integral value over the hemisphere.