Problem 299 - Calculus Volume 3

Question

[T] Evaluate \(\iint_{S}(z+y) d S,\) where \(S\) is the part of the graph of \(z=\sqrt{1-x^{2}}\) in the first octant between the \(x z\) -plane and plane \(y=3\).

Step-by-Step Solution

Answer

The integral evaluates to \( 3 + \frac{9\pi}{4} \).

Step 1: Understand the Surface S

The surface \( S \) is defined by \( z = \sqrt{1-x^{2}} \). It is a circular parabola above the \( xy \)-plane within the first octant (where \( x, y, z \geq 0 \)). It extends between the \( xz \)-plane (where \( y = 0 \)) and the plane \( y = 3 \), which bounds \( S \) in the \( y \)-direction.

Step 2: Express dS Using the Surface and Parameters

To evaluate the surface integral, we will parametrize surface \( S \). Since \( z \) is a function of \( x \), and \( x \) and \( y \) are independent, use parameterization: \( \mathbf{r}(x, y) = (x, y, \sqrt{1-x^{2}}) \). The formula for \( dS \) on a surface \( z = f(x, y) \) is \( \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy \).

Step 3: Calculate Partial Derivatives and \( dS \)

Find the partial derivatives:\[ \frac{\partial z}{\partial x} = -\frac{x}{\sqrt{1-x^{2}}}, \, \frac{\partial z}{\partial y} = 0 \]. Calculate \( dS \):\[ dS = \sqrt{1 + \left( -\frac{x}{\sqrt{1-x^{2}}} \right)^2} \, dx \, dy = \sqrt{\frac{1}{1-x^{2}}} \, dx \, dy = \frac{1}{\sqrt{1-x^{2}}} \, dx \, dy \].

Step 4: Set Up the Double Integral

Now, set up the integral of \( (z + y) \, dS \):\[ \iint_{S} (z+y) \, dS = \iint_{R} \left(\sqrt{1-x^{2}} + y \right) \cdot \frac{1}{\sqrt{1-x^{2}}} \, dx \, dy \]. \( R \) is determined by \( y = 0 \) to \( y = 3 \) and \( x = 0 \) to \( x = 1 \), due to the sphere's limitation to positive \( x \) values.

Step 5: Solve the Double Integral

Simplify and integrate \( \iint_{R} (1 + \frac{y}{\sqrt{1-x^{2}}}) \, dx \, dy \). First, integrate with respect to \( x \):\[ \int_{0}^{1} \left(1 + \frac{y}{\sqrt{1-x^{2}}}\right) \, dx \]. This computes to \( [x + y\sin^{-1}(x)]_{0}^{1} \), which simplifies to \( 1 + \frac{\pi}{2}y \). Next, integrate with respect to \( y \) from 0 to 3: \[ \int_{0}^{3} (1 + \frac{\pi}{2}y) \, dy = [y + \frac{\pi}{4}y^{2}]_{0}^{3} = 3 + \frac{9\pi}{4} \].

Key Concepts

Parametrization

Parametrization is a method used to represent a surface with two parameters, usually denoted by coordinates like \(x\) and \(y\). In this exercise, the surface \(S\) is part of a circular parabola defined by \(z = \sqrt{1-x^2}\). It is crucial to parametrize the surface because it helps us express points on the surface in terms of variables we can manipulate easily in equations.
For the surface \(S\), a point can be represented as \(\mathbf{r}(x, y) = (x, y, \sqrt{1-x^{2}})\). This parametrization describes how \(z\) changes with \(x\) and \(y\), respecting the surface's constraints.
This step simplifies assessing changes in surface direction and calculating integrals over that surface.

Partial Derivatives

Partial derivatives involve taking the derivative of a function of multiple variables with respect to one variable, holding the others constant. They are critical when dealing with surfaces like \(S\) because they show how the surface changes in each direction.
For \(z = \sqrt{1-x^2}\), we find partial derivatives:

• With respect to \(x\): \\(\frac{\partial z}{\partial x} = -\frac{x}{\sqrt{1-x^{2}}}\)
• With respect to \(y\): \\(\frac{\partial z}{\partial y} = 0\)

The partial derivative with respect to \(x\) shows how \(z\) changes as \(x\) varies, while with respect to \(y\), it indicates no change because \(z\) doesn't depend on \(y\) explicitly.
These derivatives are essential for calculating the area element \(dS\) needed in surface integrals.

Double Integral

Double integrals handle functions across two variables over a specified region, ideal for calculating quantities like area, volume, or mass over surfaces.
In this exercise, we set up a double integral for \(\iint_{S} (z+y) \, dS\). This involves the region \(R\) where \(0 \leq y \leq 3\) and \(0 \leq x \leq 1\).
The double integral \(\iint_{R} (1 + \frac{y}{\sqrt{1-x^{2}}}) \, dx \, dy\) breaks down into:

• Integrating with respect to \(x\): \(\int_{0}^{1} \left(1 + \frac{y}{\sqrt{1-x^{2}}}\right) \, dx\)
• Then with respect to \(y\): \(\int_{0}^{3} (1 + \frac{\pi}{2}y) \, dy\)

Double integrals in this form are excellent for assessing properties over defined regions, effectively evaluating the surface integral of complex surfaces.

Circular Parabola

A circular parabola is a special type of surface which can be visualized as a part of a 3D shape like a bowl or dome. In this exercise, the circular parabola is defined by the equation \(z = \sqrt{1-x^2}\).
It is bounded within the first octant, meaning the surface extends in the positive \(x, y,\) and \(z\) directions only. This constraint ensures the study of the surface is limited to practical regions of interest, crucial for interpreting results in a bounded and meaningful physical context.
The circular nature comes from the term \(\sqrt{1-x^2}\), forming a semicircle when \(z=0\), making it part of a circle rotated around the \(z\) axis, creating a hemispherical shape.