Problem 20 - Calculus Early Transcendentals

Question

Set up, but do not evaluate, an iterated integral equal to the given surface integral by projecting \(\sigma\) on (a) the \(x y\) -plane, (b) the \(y z\) -plane, and ( \(c\) ) the \(x z\) -plane. \(\iint_{\sigma} x z d S,\) where \(\sigma\) is the portion of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) in the first octant.

Step-by-Step Solution

Answer

The iterated integrals for each projection are set but not evaluated.

Step 1: Define the Surface and Iterated Integral in General Form

The surface \(\sigma\) is part of the sphere given by \(x^2 + y^2 + z^2 = a^2\), restricted to the first octant where \(x, y, ext{and} \, z \geq 0\). The surface integral is \(\iint_{\sigma} xz \, dS\). To express \(dS\) correctly when projecting on coordinate planes, parameterize the surface.

Step 2: Projection on the xy-plane

First, project \(\sigma\) onto the \(xy\)-plane. The projection equation becomes \(x^2 + y^2 \leq a^2\). Solve for \(z \) using \(z = \sqrt{a^2 - x^2 - y^2}\). For the iterated integral setup: \[\int_0^a \int_0^{\sqrt{a^2-x^2}} x \sqrt{a^2-x^2-y^2} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} \, dy \, dx\where \, \frac{\partial z}{\partial x} = -\frac{x}{\sqrt{a^2-x^2-y^2}}, \, \frac{\partial z}{\partial y} = -\frac{y}{\sqrt{a^2-x^2-y^2}}\]

Step 3: Determine Surface Area Element for xy-Projection

The surface area element \(dS\) in terms of the projected integral involves the normal vectors: \[dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy \,\] Substituting gives: \[dS = \frac{1}{\sqrt{a^2 - x^2 - y^2}} \, dx \, dy\\text{so the integral becomes: } \int_0^a \int_0^{\sqrt{a^2-x^2}} x \left(\sqrt{a^2 - x^2 - y^2}\right)^2 \, dy \, dx\]

Step 4: Projection on the yz-plane

Next, project \(\sigma\) onto the \(yz\)-plane. The projection equation becomes \(y^2 + z^2 \leq a^2\). Solve for \(x\) using \(x = \sqrt{a^2 - y^2 - z^2}\). The iterated integral is: \[\int_0^a \int_0^{\sqrt{a^2-y^2}} \sqrt{a^2-y^2-z^2} \, z \, \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} \, dz \, dy\]

Step 5: Determine Surface Area Element for yz-Projection

In terms of \(yz\)-projection, the surface element \(dS\) involves: \[dS = \sqrt{1 + \left(\frac{\partial x}{\partial y}\right)^2 + \left(\frac{\partial x}{\partial z}\right)^2} \, dy \, dz\= \frac{1}{\sqrt{a^2 - y^2 - z^2}} \, dy \, dz\] The iterated integral becomes: \(\int_0^a \int_0^{\sqrt{a^2-y^2}} \, \sqrt{a^2 - y^2 - z^2} \, z \, dy \, dz\)

Step 6: Projection on the xz-plane

Finally, project \(\sigma\) onto the \(xz\)-plane. The projection equation is \(x^2 + z^2 \leq a^2\). Solve for \(y\) using \(y = \sqrt{a^2 - x^2 - z^2}\). The iterated integral becomes: \[\int_0^a \int_0^{\sqrt{a^2-x^2}} x z \sqrt{1+\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2} \, dz \, dx\]

Step 7: Determine Surface Area Element for xz-Projection

The differential of surface area in xz-projection: \[dS = \sqrt{1 + \left(\frac{\partial y}{\partial x}\right)^2 + \left(\frac{\partial y}{\partial z}\right)^2} \, dz \, dx\= \frac{1}{\sqrt{a^2 - x^2 - z^2}} \, dz \, dx\]. Substitute in the iterated integral to get: \(\int_0^a \int_0^{\sqrt{a^2-x^2}} x z \frac{1}{\sqrt{a^2 - x^2 - z^2}} \, dz \, dx\)

Key Concepts

Surface Integral

A surface integral allows us to perform integration over a surface, much like a double integral does for a region in a plane. In this scenario, it involves the integration of scalar fields or vector fields over 3D surfaces. For the surface integral \( \iint_{\sigma} xz \, dS \), we integrate the product \( xz \) over the surface \( \sigma \), which is part of a sphere.

Here, \( \sigma \) is defined by the sphere equation \( x^2 + y^2 + z^2 = a^2 \), but only in the first octant where \( x, y, \text{and} \ z \geq 0 \). This means that only the portion of the surface that lies in the first eighth of the 3D space is considered.

The surface integral thus quantifies the cumulative effect (like weighing) of the term \( xz \) distributed across this spherical surface. It's crucial to understand how this integrates once it's cast into iterated integrals based on surface projections.

Projection in Coordinate Planes

Projection involves casting the surface onto one of the coordinate planes, like flattening the surface into a 2D shape. This step is helpful in simplifying the surface integral into an iterated integral, as it allows us to express the surface in coordinate terms that are easier to handle.

In this exercise, the surface \( \sigma \) is projected sequentially onto three planes:

• **The \( xy \)-plane**: By eliminating \( z \), the surface flattens to the circle \( x^2 + y^2 \leq a^2 \), considering only non-negative x and y. The integral configuration derived is \[ \int_0^a \int_0^{\sqrt{a^2-x^2}} x \left(\sqrt{a^2-x^2-y^2}\right)^2 \ dy \ dx \]
• **The \( yz \)-plane**: This projection, removing \( x \), gives \( y^2 + z^2 \leq a^2 \), with the iterated integral \[ \int_0^a \int_0^{\sqrt{a^2-y^2}} \sqrt{a^2-y^2-z^2} \ z \ dy \ dz \]
• **The \( xz \)-plane**: Removing \( y \), the equation is \( x^2 + z^2 \leq a^2 \), leading to the integral \[ \int_0^a \int_0^{\sqrt{a^2-x^2}} x z \frac{1}{\sqrt{a^2-x^2-z^2}} \ dz \ dx \]

Each projection offers a unique view of the surface, simplifying the complex 3D surface into manageable 2D components.

Surface Area Element

To calculate the surface integral, we need a surface area element \( dS \), which corresponds to a tiny patch of area on the surface. Its determination is a fundamental step when solving surface integrals since it affects how the integral accumulates values across the surface.

For each projection of the surface, the surface area element \( dS \) is determined by the gradient of the surface's parameterization in terms of the other variables:

• For **\( xy \)-projection**: The surface element \( dS \) is expressed as \( \frac{1}{\sqrt{a^2-x^2-y^2}} \, dx \, dy \).
• For **\( yz \)-projection**: The surface element becomes \( \frac{1}{\sqrt{a^2-y^2-z^2}} \, dy \, dz \).
• For **\( xz \)-projection**: It's \( \frac{1}{\sqrt{a^2-x^2-z^2}} \, dz \, dx \).

These expressions result from evaluating the normal vector's length in each projection, factoring in the implicit function defining the surface. Such steps are critical as they enable transformation of the surface integrals into iterated integrals, simplifying the overall computation.