Problem 2 - Advanced Engineering Mathematics

Question

Find the surface area of that portion of the plane \(2 x+3 y+4 z=12\) that is above the region in the first quadrant bounded by the graph \(r=\sin 2 \theta\).

Step-by-Step Solution

Answer

The surface area is \(\frac{\pi \sqrt{29}}{32}\).

Step 1: Convert Plane to Function of z

The plane is given by the equation \(2x + 3y + 4z = 12\). We need to express \(z\) in terms of \(x\) and \(y\) to find the surface area. Solving for \(z\), we have \(z = 3 - \frac{1}{2}x - \frac{3}{4}y\).

Step 2: Determine the Region of Integration

The region in the first quadrant is bounded by \(r = \sin 2 \theta\). In polar coordinates, this is the region for \(0 \leq r \leq \sin 2\theta\) and \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 3: Calculate the Surface Element

The surface area is calculated using the integral of the surface element. The formula is \(dS = \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}\, dx\, dy\). For our function \(z(x,y) = 3 - \frac{1}{2}x - \frac{3}{4}y\), the partial derivatives are:\[ \frac{\partial z}{\partial x} = -\frac{1}{2}, \quad \frac{\partial z}{\partial y} = -\frac{3}{4}. \]Substitute these into the equation for \(dS\) to get: \[dS = \sqrt{1 + \left(-\frac{1}{2}\right)^2 + \left(-\frac{3}{4}\right)^2}\, dx\, dy = \sqrt{1 + \frac{1}{4} + \frac{9}{16}}\, dx\, dy = \sqrt{\frac{29}{16}}\, dx\, dy.\]

Step 4: Set Up the Integral for Surface Area

Convert the integral to polar coordinates, where \(x = r \cos \theta\) and \(y = r \sin \theta\). The surface element \(dx\, dy\) becomes \(r\, dr\, d\theta\):\[\iint_{R} \sqrt{\frac{29}{16}}\, dx\, dy = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin 2\theta} \sqrt{\frac{29}{16}}\, r\, dr\, d\theta.\]

Step 5: Evaluate the Integral

Evaluate the internal integral with respect to \(r\):\[\int_{0}^{\sin 2\theta} r\, dr = \left[ \frac{r^2}{2} \right]_{0}^{\sin 2\theta} = \frac{1}{2} (\sin^2 2\theta).\]Substitute back and evaluate the external integral with respect to \(\theta\):\[\int_{0}^{\frac{\pi}{2}} \frac{1}{2} \sin^2 2\theta \sqrt{\frac{29}{16}}\, d\theta.\]The integral simplifies to:\[\frac{\sqrt{29}}{8} \int_{0}^{\frac{\pi}{2}} \sin^2 2\theta\, d\theta.\]Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), the integral becomes:\[\frac{\sqrt{29}}{8} \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 4\theta}{2}\, d\theta = \frac{\sqrt{29}}{16} \left[ \theta - \frac{1}{4} \sin 4\theta \right]_{0}^{\frac{\pi}{2}}.\]The evaluated definite integral is:\[\frac{\sqrt{29}}{16} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi \sqrt{29}}{32}.\]

Step 6: Conclusion

The surface area of the portion of the plane above the specified region is \(\frac{\pi \sqrt{29}}{32}.\).

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system used to locate a point in a plane. Unlike Cartesian coordinates which use the horizontal and vertical measurements (x and y), polar coordinates use a distance and an angle to pinpoint a location. The angle is measured from a fixed direction, usually the positive x-axis, and the distance is measured from a fixed point called the pole (analogous to the origin in Cartesian coordinates).

In the given exercise, we deal with a region in the first quadrant described by the polar equation \( r = \sin 2\theta \). This equation defines a boundary for the region we are considering. Here, \( r \) represents the distance from the origin, and \( \theta \) is the angle ranging from 0 to \( \frac{\pi}{2} \). Within this boundary, the values of \( r \) vary from 0 to \( \sin 2\theta \).

Using polar coordinates in integration simplifies the process when dealing with circular or radial symmetry, as seen in this exercise.

Partial Derivatives

Partial derivatives are tools from calculus used to study functions with more than one variable. When you take a derivative of a function with respect to one variable while keeping other variables constant, you're finding the partial derivative.

In the exercise, the plane equation \( z = 3 - \frac{1}{2}x - \frac{3}{4}y \) is expressed as a function of \( x \) and \( y \). To compute the surface area of the plane over the given region, partial derivatives with respect to \( x \) and \( y \) are needed.

• The partial derivative with respect to \( x \) is given by \( \frac{\partial z}{\partial x} = -\frac{1}{2} \)
• The partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = -\frac{3}{4} \)

These derivatives help in determining the slope of the plane in the x and y directions, which are necessary to compute the element of surface area \( dS \). Understanding partial derivatives is crucial for analyzing how changes in one variable affect the function.

Integration

Integration is the process of calculating the area under a curve, or more generally, the accumulation of quantities. In the context of surface area calculation, integration helps in summing up infinitesimal elements of area to find the total surface area.

For the exercise, we converted the given equation of the plane into an integral in polar coordinates over a specific region. The expression for the surface element was derived using integration:

• The surface area element \( dS \) was expressed as \( \sqrt{\frac{29}{16}} \, dx \, dy \)
• In polar coordinates, this converts the element \( dx \, dy \) to \( r \, dr \, d\theta \)

The integral is set up as \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sin 2\theta} \sqrt{\frac{29}{16}} r \ dr \ d\theta \).

This integration over \( r \) and \( \theta \) captures the entire area as defined by the function \( r = \sin 2\theta \). Integration is the key process in calculus that permits the evaluation of such areas or volumes in multi-dimensional spaces.

Plane Equations

A plane equation is used to describe a flat, two-dimensional surface in a three-dimensional space. This equation typically has the form \( Ax + By + Cz = D \), where \( A \), \( B \), and \( C \) are coefficients that determine the orientation of the plane in space.

In the problem, the equation \( 2x + 3y + 4z = 12 \) describes the plane we are interested in. To calculate the surface area of a portion of this plane, we begin by expressing \( z \) as a function of \( x \) and \( y \) (i.e., \( z = 3 - \frac{1}{2}x - \frac{3}{4}y \)).

This transformation is necessary for applying partial derivatives and simplifying the integration process to calculate the surface area as we move across the defined region. Plane equations essentially describe the relationships between x, y, and z on the surface of a plane, aiding us in finding solutions to such geometric problems.