Problem 3 - Calculus Volume 3

Question

In the following exercises, estimate the volume of the solid under the surface \(z=f(x, y)\) and above the rectangular region \(R\) by using a Riemann sum with \(m=n=2\) and the sample points to be the lower left corners of the subrectangles of the partition. $$ f(x, y)=\sin x-\cos y, \quad R=[0, \pi] \times[0, \pi] $$

Step-by-Step Solution

Answer

. V \approx \frac{\pi^2}{2} - \frac{\pi^2}{2} = 0

Step 1: Divide the Region

The region \(R = [0, \pi] \times [0, \pi]\) is to be divided into \(m \times n\) subrectangles. Given \(m = 2\) and \(n = 2\), \(R\) is divided into 4 subrectangles.

Step 2: Calculate Dimensions of Each Subrectangle

Each dimension of the subrectangles can be calculated as \(\Delta x = \frac{\pi - 0}{2} = \frac{\pi}{2}\) and \(\Delta y = \frac{\pi - 0}{2} = \frac{\pi}{2}\).

Step 3: Determine Sample Points

The sample points for each subrectangle are the lower left corners. The points are: \((0, 0), (0, \frac{\pi}{2}), (\frac{\pi}{2}, 0), (\frac{\pi}{2}, \frac{\pi}{2})\).

Step 4: Evaluate the Function at Sample Points

Calculate the function value at each sample point: - At \((0, 0)\), \(f(0,0) = \sin(0) - \cos(0) = -1\). - At \((0, \frac{\pi}{2})\), \(f(0,\frac{\pi}{2}) = \sin(0) - \cos(\frac{\pi}{2}) = 0\). - At \((\frac{\pi}{2}, 0)\), \(f(\frac{\pi}{2},0) = \sin(\frac{\pi}{2}) - \cos(0) = 1 - 1 = 0\). - At \((\frac{\pi}{2}, \frac{\pi}{2})\), \(f(\frac{\pi}{2},\frac{\pi}{2}) = \sin(\frac{\pi}{2}) - \cos(\frac{\pi}{2}) = 1\).

Step 5: Compute the Riemann Sum

The volume is estimated by the sum: \[ V \approx \sum_{i=1}^{4} f(x_i, y_i) \Delta x \Delta y = (-1) \left(\frac{\pi}{2}\right)^2 + (0) \left(\frac{\pi}{2}\right)^2 + (0) \left(\frac{\pi}{2}\right)^2 + (1) \left(\frac{\pi}{2}\right)^2 \] This simplifies to: \[ V \approx \left((-1) + 1\right) \left(\frac{\pi}{2}\right)^2 = 0 \]}],

Step 6: Description

The estimated volume under the surface and above the region is \(0\).

Step 7: Solution Steps

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Step 8: Short Answer

The estimated volume is 0.

Step 9: Solution Steps

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Step 10: Solution Steps

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Key Concepts

Volume Estimation

Volume estimation in mathematics involves calculating the space occupied by a three-dimensional object. In multivariable calculus, this process often requires methods such as Riemann sums, especially when dealing with regions under surfaces represented by functions like \(z = f(x, y)\). Here, the surface is above a finite rectangular region, and our task is to estimate the volume using discrete sample points.

To estimate the volume, we divide the region of interest into subregions or subrectangles. In our example, we divide \(R = [0, \pi] \times [0, \pi]\) into four equal parts. By calculating the function values at these points, and multiplying them by the dimensions of the subrectangles, we get an estimate of the volume. This method is handy when integrals are difficult to compute directly.

Multivariable Calculus

Multivariable calculus extends single-variable calculus to functions of two or more variables. It builds on the basic principles of calculus, like differentiation and integration, but applied to functions \(f(x, y)\) or \(f(x, y, z)\). In our case, the surface \(z = f(x, y) = \sin x - \cos y\) covers a region in the \(xy\)-plane.

This type of calculus introduces concepts like partial derivatives, which measure how a function changes with just one of its variables, while keeping others constant. Similarly, it uses multiple integrals to compute volumes or areas and when estimating volumes under surfaces, techniques like Riemann sums offer approximations using simple geometric shapes.

Sample Points Calculation

Sample points are specific points in each subrectangle or subregion that help in calculating Riemann sums. These points are where we evaluate the function to approximate the sum.

For our exercise, the sample points are chosen to be the lower-left corners of the subrectangles. Each of these subrectangles is formed by dividing the region \(R = [0, \pi] \times [0, \pi]\) into four by lines parallel to the axes. In this way, you get sample points \((0, 0), (0, \frac{\pi}{2}), (\frac{\pi}{2}, 0), (\frac{\pi}{2}, \frac{\pi}{2})\).

Using these sample points, we evaluate the function values and further use these to find the Riemann sum, which leads to our volume estimation. Choosing where to evaluate the function is crucial, as it directly affects the estimation outcome.

Integral Approximation

Integral approximation is the process of finding an approximate value for a definite integral using numerical methods. It becomes particularly useful when an integral is difficult or impossible to solve analytically.

Riemann sums provide one such method. By summing the area of small rectangles, we can approximate the integral of a function over a region. In this method, the choice of sample points significantly influences the sum. In our example, each sample point's height is determined by evaluating \(f(x, y) = \sin x - \cos y\) at the specific point.

The width and height of these rectangles correspond to \(\Delta x\) and \(\Delta y\), respectively. Multiplying them with the function's value gives the volume under the curve for that small rectangle. Summing these volumes results in an approximation of the total volume under the curve. This approach simplifies complex integration tasks into manageable calculations.