Question

Set up the iterated integral that computes the surface area of the given surface over the region \(R .\) \(f(x, y)=\sin x \cos y ; \quad R\) is the rectangle with bounds \(0 \leq\) \(x \leq 2 \pi, \quad 0 \leq y \leq 2 \pi\).

Short answer

\(\int_0^{2\pi} 2\pi \sqrt{1 + \cos^2 y} \, dy\).

Step-by-step solution

Understand the Surface Area Formula

The surface area of a surface defined by a function \(f(x, y)\) over a region \(R\) is given by the integral \( \iint_R \sqrt{1 + (f_x)^2 + (f_y)^2} \, dA \), where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\) respectively.

Compute the Partial Derivatives

Calculate the partial derivative of \(f(x, y) = \sin x \cos y\) with respect to \(x\): \(f_x = \cos x \cos y\). Next, calculate the partial derivative with respect to \(y\): \(f_y = -\sin x \sin y\).

Substitute into Surface Area Formula

Substitute \(f_x\) and \(f_y\) into the surface area formula. The expression under the square root becomes \(1 + \cos^2 x \cos^2 y + \sin^2 x \sin^2 y\).

Simplify the Expression

Use the trigonometric identity \(\cos^2 A + \sin^2 A = 1\) to simplify the expression: \(1 + \cos^2 x \cos^2 y + \sin^2 x \sin^2 y = 1 + \cos^2 y\). Thus, the integrand is \(\sqrt{1 + \cos^2 y}\).

Set Up the Iterated Integral

Set the limits of integration over the region \(R\) which is defined by the bounds \([0, 2\pi]\) for both \(x\) and \(y\). The iterated integral becomes \( \int_0^{2\pi} \int_0^{2\pi} \sqrt{1 + \cos^2 y} \, dx \, dy \).

Evaluate the Iterated Integral

Keep \(\sqrt{1 + \cos^2 y}\) outside the inner integral as it doesn't depend on \(x\). Solve the inner integral \(\int_0^{2\pi} dx = 2\pi\), and then the iterated integral becomes \( \int_0^{2\pi} 2\pi \sqrt{1 + \cos^2 y} \, dy \).

Key concepts

Surface Area

When finding the surface area of a surface defined by a function of two variables, like \(f(x, y)\), over a specified region \(R\), you need to set up a special kind of integral. This is known as an iterated integral for surface area. The general formula for the surface area is: \\[ \iint_R \sqrt{1 + (f_x)^2 + (f_y)^2} \, dA \] \where \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.

• The square root term represents an adjustment for how the surface moves in 3D space.
• This formula helps relate the flat 2D region \(R\) to the curved surface defined by \(f\).

This integral will give you the total surface area when evaluated over the region \(R\). Here, we observe how the curve \(\sin x \cos y\) expands in a wave-like pattern and by calculating its surface area, we account for the undulations over the rectangle \([0, 2\pi]\) for both \(x\) and \(y\).
Calculate the integral to find out the exact surface area of this surface exposed over the defined region.

Partial Derivatives

Partial derivatives are crucial in calculating surface areas of functions of more than one independent variable, like \(f(x, y)\). A partial derivative allows us to see how the function changes with respect to one variable while holding the other constant.
\(f_x\) denotes the partial derivative with respect to \(x\) and is calculated by differentiating \(f(x, y)\) treating \(y\) as a constant. Similarly, \(f_y\) is gained by differentiating \(f(x, y)\) with \(x\) as a constant.
For \(f(x, y) = \sin x \cos y\):

• \(f_x = \cos x \cos y\): Here, the derivative of \(\sin x\) with respect to \(x\) is \(\cos x\) while \(\cos y\) remains as a multiplicative constant.
• \(f_y = -\sin x \sin y\): Upon differentiation with respect to \(y\), the derivative of \(\cos y\) is \(-\sin y\), and \(\sin x\) remains as is.

These derivatives help adjust the original surface area formula to account for the local slopes or steepness along both axes, thus improving the accuracy of surface area calculations.

Trigonometric Identities

Trigonometric identities play a huge role in simplifying complex expressions, especially when integrating functions that contain trigonometric functions like \(\sin\) and \(\cos\).
One of the fundamental identities used frequently is:\[\cos^2 A + \sin^2 A = 1\] This relation allows us to reduce expressions significantly by converting squared trigonometric terms into more manageable parts.
In the original exercise, this identity transforms the expression: \\[1 + \cos^2 x \cos^2 y + \sin^2 x \sin^2 y\] \into \\[1 + \cos^2 y\] \since \(\cos^2 x + \sin^2 x = 1\).

• Using these identities simplifies the integrand, making the integration process more straightforward and manageable.
• By reducing complex functions, these identities help in reducing computational complexities.

Identities like these underpin much of trigonometry, helping solve problems that may initially seem very dense or complex.