Problem 298 - Calculus Volume 3

Question

[T] Evaluate \(\iint_{S}(x z / y) d S,\) where \(S\) is the portion of cylinder \(x=y^{2}\) that lies in the first octant between planes \(z=0, z=5, y=1,\) and \(y=4\).

Step-by-Step Solution

Answer

Integrate with respect to \(z\) and then \(y\) to find that the integral evaluates to approximately 708.33.

Step 1: Understand the Surface S

The surface \(S\) is a portion of the cylinder defined by the equation \(x = y^2\) lying in the first octant. It is bounded by the planes \(z = 0\), \(z = 5\), \(y = 1\), and \(y = 4\). This means \(S\) is a curved surface segment that extends from \(y = 1\) to \(y = 4\) along the \(x = y^2\) curve, and from \(z = 0\) to \(z = 5\).

Step 2: Parametrize the Surface

We can parametrize the surface \(S\) using parameters \(y\) and \(z\). Let \(x = y^2\), \(y\) varies from 1 to 4, and \(z\) varies from 0 to 5. Thus, the parametrization is \(\mathbf{r}(y, z) = (y^2, y, z)\).

Step 3: Compute the Surface Normal Vector

Compute the cross product of the partial derivatives of the parametrization: \(\mathbf{r}_y = (2y, 1, 0)\) and \(\mathbf{r}_z = (0, 0, 1)\). The normal vector is \(\mathbf{n} = \mathbf{r}_y \times \mathbf{r}_z = (1, -2y, 2y)\).

Step 4: Express the Integrand in Terms of Parameters

The integrand \(\frac{xz}{y}\) becomes \(\frac{y^2z}{y} = yz\) after substituting \(x = y^2\).

Step 5: Set Up the Surface Integral

The integral is \(\iint_S yz \, dS = \int_1^4 \int_0^5 yz \sqrt{1^2 + (-2y)^2 + (2y)^2} \, dz \, dy\). Simplify the square root to \(\sqrt{1 + 4y^2 + 4y^2} = \sqrt{1 + 8y^2}\).

Step 6: Evaluate the Integral

First, solve the integral with respect to \(z\): \[\int_0^5 yz \, dz = y \left[\frac{z^2}{2} \right]_0^5 = \frac{25y}{2}\].Integrate with respect to \(y\): \[\int_1^4 \frac{25y}{2} \sqrt{1 + 8y^2} \, dy\].This requires substitution: \(u = 1 + 8y^2\), \(du = 16y \, dy\), so \(dy = \frac{du}{16y}\). Solve this integral for a final result.

Step 7: Compute \\(\int u^{1/2} du\\)

Perform the substitution \(u = 1 + 8y^2\), thus the integral becomes \[\int (1/16) \times \frac{25y}{2} u^{1/2} \, du = \frac{25}{32} \int u^{1/2} \, du\].The integral results in \(\frac{25}{32} \times \frac{2}{3} (u^{3/2}) = \frac{25}{48} (1 + 8y^2)^{3/2}\).

Step 8: Evaluate From 1 to 4

Evaluate the expression from Step 7 between \(y = 1\) and \(y = 4\):\[\frac{25}{48} \left((1 + 8(4)^2)^{3/2} - (1 + 8(1)^2)^{3/2}\right)\].Calculate \((1 + 128)^{3/2} - (1 + 8)^{3/2}\) which gives the final result of the integral.

Key Concepts

Parametrization

Parametrization is a technique used to express a surface in terms of two independent parameters, typically denoted as \( u \) and \( v \). This process helps to represent complex surfaces in a navigable manner. In our exercise, the surface \( S \) is part of a cylinder defined by the equation \( x = y^2 \). Here, we chose "\(y\)" and "\(z\)" as the parameters that vary, making the process simpler.

The parametric equations become \( x = y^2, y = y, \) and \( z = z \). This corresponds to the vector function \( \mathbf{r}(y, z) = (y^2, y, z) \). With this parametrization, the parameter \( y \) can range from 1 to 4, and \( z \) can range from 0 to 5. This choice of parameters reveals the extent and orientation of the portion of the cylinder involved in the integral calculation.

• Parametrization is key for converting a surface into a format that simplifies integration.
• It helps manage the constraints on a surface specified in 'xyz' space.

Cylinder

Cylinders are basic geometric shapes that appear frequently in calculus problems, particularly those involving surface integrals. A cylinder in three-dimensional space can be imagined as an extended circular shape that varies along the axis of projection.

In this exercise, the cylinder equation is \( x = y^2 \). This defines a parabola rotated along the axis perpendicular to the y-z plane, rather than a traditional circular cylinder. The section of interest lies in the first octant, which means all coordinates \( x, y, \) and \( z \) must be positive.

• The bounds of the cylinder segment are set by the constraints: \( y = 1 \) to \( y = 4 \) and \( z = 0 \) to \( z = 5 \).
• Understanding the geometric shape of the cylinder helps in visualizing the region to integrate over.

Normal Vector

A normal vector is essential in surface integrals as it indicates the direction perpendicular to the surface at any point on the surface. This helps in determining the differential area element \( dS \) in the integral calculation.

To find this for our parametrized surface, we take the cross product of the partial derivatives of the parametrization vector \( \mathbf{r}(y, z) \). The calculations yield:- Partial derivative with respect to \( y \), \( \mathbf{r}_y = (2y, 1, 0) \)- Partial derivative with respect to \( z \), \( \mathbf{r}_z = (0, 0, 1) \)- The normal vector, \( \mathbf{n} = \mathbf{r}_y \times \mathbf{r}_z = (1, -2y, 2y) \).

• The normal vector is crucial for expressing \( dS \) in terms of \( dy \) and \( dz \).
• It integrates both directional properties and the magnitude of the surface area element.

Substitution Method

The substitution method is a powerful tool in calculus that simplifies integration by changing variables. This method allows a complicated integral to be transformed into an easier one, often by letting \( u = g(y) \) and calculating its derivative.

In the given problem, substitution helps in integrating the function with respect to \( y \). Here the integrand \( yz \sqrt{1 + 8y^2} \) becomes more manageable with a substitution such as \( u = 1 + 8y^2 \), leading to \( du = 16y \, dy \). Consequently, \( dy \) is replaced by \( \frac{du}{16y} \) making the integral \( \int u^{1/2} \, du \) solvable.

• The method leads to simplified functions that are easier to integrate.
• It requires understanding the derivative and integral relationship thoroughly.