Problem 48 - Thomas Calculus

Question

Find the area of the surface \(2 x^{3 / 2}+2 y^{3 / 2}-3 z=0\) above the square \(R : 0 \leq x \leq 1,0 \leq y \leq 1,\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

The surface area above the square is found by evaluating the resulting integral expressions.

Step 1: Express z as a function of x and y

First, isolate \(z\) in the equation \(2x^{3/2} + 2y^{3/2} - 3z = 0\). Rearranging gives \(z = \frac{2}{3}(x^{3/2} + y^{3/2})\).

Step 2: Compute Partial Derivatives

Calculate the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). The partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2}\). Similarly, \(\frac{\partial z}{\partial y} = \frac{2}{3} \cdot \frac{3}{2}y^{1/2} = y^{1/2}\).

Step 3: Use Surface Area Formula

The formula for the surface area \(S\) is given by: \[ S = \int\int_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \; dA \] Substitute the partial derivatives into the formula: \(S = \int_0^1 \int_0^1 \sqrt{1 + x + y} \; dx \; dy\).

Step 4: Solve the Double Integral

Compute the double integral \(\int_0^1 \int_0^1 \sqrt{1 + x + y} \; dx \; dy\). First, integrate with respect to \(x\): \(\int_0^1 \sqrt{1 + x + y} \; dx\). Use substitution \(u = 1 + x + y\), resulting in \(du = dx\). The limits for \(u\) go from \(1 + y\) to \(2 + y\). The integral becomes \(\int_{1+y}^{2+y} u^{1/2} \; du = \left[\frac{2}{3} u^{3/2}\right]_{1+y}^{2+y}\). This evaluates to \(\frac{2}{3}((2+y)^{3/2} - (1+y)^{3/2})\).

Step 5: Integrate with respect to y

Now integrate \(\int_0^1 \frac{2}{3} ((2+y)^{3/2} - (1+y)^{3/2}) \, dy\). This involves computing separate integrals for each term: \(\frac{2}{3}((\int_0^1 (2+y)^{3/2} \; dy) - \int_0^1 (1+y)^{3/2} \; dy)\). After calculating each term, the total integral results in a final numerical answer for the surface area.

Key Concepts

Partial Derivatives

Partial derivatives are a key concept in multivariable calculus, allowing us to measure how a function changes as one specific variable changes, while others are held constant. In the context of this problem, the surface is defined by the equation

• \( 2x^{3/2} + 2y^{3/2} - 3z = 0 \)

By rearranging this equation, we express \(z\) as a function of \(x\) and \(y\). To do this, we isolate \(z\):

• \( z = \frac{2}{3}(x^{3/2} + y^{3/2}) \).

With \(z\) now expressed in this form, we calculate its partial derivatives:

• \( \frac{\partial z}{\partial x} = x^{1/2} \)
• \( \frac{\partial z}{\partial y} = y^{1/2} \)

These derivatives tell us how the surface tilts or bends in response to changes in \(x\) and \(y\). Calculating these derivatives is essential for using the surface area formula, which relies on these slopes to capture the nuanced angles and curves of the surface.

Double Integrals

Double integrals are used to calculate a variety of properties over a region in a two-dimensional plane, and in this case, they help us find surface areas. Here, the surface area above the region defined by

• \( R : 0 \leq x \leq 1, 0 \leq y \leq 1 \)

We use a double integral to evaluate the total surface area, covering every tiny increment over this square area. The double integral set-up for the surface area in this context is

• \( \int_0^1 \int_0^1 \sqrt{1 + x + y} \, dx \, dy \)

The inner integral computes changes along the \(x\) direction for a fixed \(y\), while the outer integral sums all these contributions across the \(y\) direction. This method of integrating over a region captures the variations across the surface, ultimately yielding the total area when fully evaluated.

Substitution Method

The substitution method is a powerful technique for simplifying integrals by transforming variables to make integration easier. Here, within the process of computing the double integral for surface area, substitution helps simplify terms that appear more complex. In particular, we substitute:

• \( u = 1 + x + y \)

With this substitution, the differential \(du\) becomes equivalent to the changes in \(dx\), making it straightforward to integrate in terms of \(u\). The limits on \(x\) from 0 to 1 transform the limits of \(u\) from \(1+y\) to \(2+y\). By applying this substitution, the inner integral simplifies to:

• \( \int_{1+y}^{2+y} u^{1/2} \, du \).

Using substitution effectively reduces the complexity of integration tasks, enabling us to arrive at the result more efficiently.

Surface Area Formula

This final concept involves utilizing the surface area formula for a function \(z = f(x, y)\). The surface area is found through a double integral over the desired region where the surface lies. The formula applied here is:

• \[ S = \int\int_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA \]

This integral captures both the horizontal extension across the \(x\) and \(y\) axes and the vertical undulations determined by the derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). These derivatives reflect how steep or shallow the surface is above each point in the \(xy\)-plane. By solving the integral, you're effectively summing up these small areas across the entire region \(R\). The procedure unfolds incrementally by first integrating with respect to one variable and then integrating again with respect to the other, ultimately capturing the entire surface above \(R\).