Question

Find the area of the given surface over the region \(R\). \(f(x, y)=3 x-7 y+2 ; R\) is the rectangle with opposite corners (-1,0) and (1,3).

Short answer

The area of the surface over the region is 0.

Step-by-step solution

Identify the Surface Function and Region

The surface function is given by \(f(x, y) = 3x - 7y + 2\). The region \(R\) is a rectangle with opposite corners at \((-1, 0)\) and \((1, 3)\). This means the rectangle is defined by \(-1 \leq x \leq 1\) and \(0 \leq y \leq 3\).

Determine the Bounds for Integration

To find the area under the surface over the region \(R\), we will set up the double integral. The region \(R\) provides the bounds for the integral: \(-1 \leq x \leq 1\) and \(0 \leq y \leq 3\).

Set Up the Double Integral

The double integral of the function \(f(x, y)\) over the region \(R\) is given by:\[ \int_{x=-1}^{x=1} \int_{y=0}^{y=3} (3x - 7y + 2) \, dy \, dx \]

Evaluate the Inner Integral

Evaluate the inner integral with respect to \(y\):\[ \int_{0}^{3} (3x - 7y + 2) \, dy = [3xy - \frac{7}{2}y^2 + 2y]_{0}^{3} \]Substituting the limits of \(y\), we get:\[ [3x(3) - \frac{7}{2}(3)^2 + 2(3)] - [3x(0) - \frac{7}{2}(0)^2 + 2(0)] = 9x - \frac{63}{2} + 6 \]which simplifies to \(9x - \frac{51}{2}\).

Evaluate the Outer Integral

Now evaluate the outer integral with respect to \(x\):\[ \int_{-1}^{1} (9x - \frac{51}{2}) \, dx = [\frac{9}{2}x^2 - \frac{51}{2}x]_{-1}^{1} \]Substituting the limits for \(x\), we have:\[ [\frac{9}{2}(1)^2 - \frac{51}{2}(1)] - [\frac{9}{2}(-1)^2 - \frac{51}{2}(-1)] \]This simplifies to \((-\frac{42}{2}) - (\frac{42}{2}) = -21 + 21 = 0\).

Conclude the Calculation

The calculated value of the double integral over the region \(R\), which represents the area of the surface over this region, is 0. Since this is a plane in three dimensions, we interpret the integral as calculating the net signed area under the surface, which is zero.

Key concepts

Double Integral

The double integral is a powerful tool in multivariable calculus. It helps us calculate the volume under a surface over a specified region in a 2-dimensional space. Imagine you have a wavy, bumpy area, and you want to calculate how much space is beneath it when stretched out. That's where a double integral comes in.

• A double integral sums up infinitely many tiny areas under a surface, like adding up a lot of small rectangles.
• In our exercise, the function defines the surface, and we calculate over the region in the xy-plane.

For the given surface function, we set up the double integral to find the net area between the surface and the xy-plane. The result of the double integral comes from calculating two layers of integration — first with respect to one variable, then with the other. It can be seen like peeling an onion, layer by layer.

Surface Area

When dealing with functions that describe surfaces, like the function in our problem, calculating the surface area involves understanding how the surface stretches over the region. The function given is a plane in a 3D space formed by the algebraic equation. Its surface can have some positive and some negative areas relative to the xy-plane, which cancel each other out when calculating net area.

• The surface' height depends on the values of the function, like peaks and valleys in terrain.
• When we calculate the integral, we're essentially summing up these rises and falls.

In this particular problem, the surface equilibrium over the region results in a total area of zero, highlighting how some areas above the plane are exactly balanced by those below.

Region of Integration

To find the area under a surface, we have to specify the region over which we calculate the double integral. The region in our case is clearly defined by a rectangle with corners at (-1,0) and (1,3). This rectangle serves as our playground, where the integration dance takes place.

• The bounds define the minimum and maximum values for each variable.
• In a rectangular region like this one, boundaries are straight lines parallel to the axis, simplifying the setup.

The region of integration is the domain for our calculations: where we decide which parts of the surface to include in the sum. Understanding the limits simplifies the integral setup process, ensuring that we only calculate over the desired area.