Question

Find the surface area \(S\) of the given surface \(\mathcal{S}\). (The associated integrals are computable without the assistance of technology.) \(\mathcal{S}\) is the plane \(z=x+2 y\) over the triangle with vertices at (0,0),(1,0) and (0,1).

Short answer

The surface area is \( \frac{\sqrt{6}}{2} \).

Step-by-step solution

Identify the region of integration

The triangle for the region of integration is formed by the vertices (0,0), (1,0), and (0,1). In the coordinate plane, you can see this forms a right triangle with the line connecting (1,0) and (0,1) given by the equation \(y = 1 - x\).

Determine the parameters for the plane

The plane equation is given by \(z = x + 2y\). To find the surface area of the plane over the given region, you will use surface integrals and need to parameterize the plane. However, this form is already suitable for the application of surface integrals directly.

Set up the surface area integral formula

The surface area \(S\) of a plane \(z = f(x, y)\) over a region \(R\) is given by the integral \(S = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA\). Here, \(f(x, y) = x + 2y\).

Calculate the partial derivatives

Compute \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). You find that \(\frac{\partial f}{\partial x} = 1\) and \(\frac{\partial f}{\partial y} = 2\).

Substitute derivatives in the formula

Substitute the partial derivatives into the surface area integral: \(S = \iint_R \sqrt{1 + 1^2 + 2^2} \, dA = \iint_R \sqrt{1 + 1 + 4} \, dA = \iint_R \sqrt{6} \, dA\).

Express the double integral over the region

The region \(R\) is the triangle in the xy-plane bounded by \(0 \leq x \leq 1\) and \(0 \leq y \leq 1-x\). Therefore, the integral becomes \( S = \int_0^1 \int_0^{1-x} \sqrt{6} \, dy \, dx \).

Perform the integration

Evaluate the inner integral \( \int_0^{1-x} \sqrt{6} \, dy = \sqrt{6}y \bigg|_0^{1-x} = \sqrt{6}(1-x)\). Now, evaluate the outer integral \( \int_0^1 \sqrt{6}(1-x) \, dx\). This gives \( \sqrt{6} \left(x - \frac{x^2}{2}\right) \bigg|_0^1 = \sqrt{6} \left(1 - \frac{1}{2}\right) = \frac{\sqrt{6}}{2} \).

Find the surface area

Hence, the surface area of the plane over the given triangular region is \( \frac{\sqrt{6}}{2} \).

Key concepts

Surface Area

When dealing with surfaces in multivariable calculus, understanding how to calculate the surface area is essential. For the plane given in this scenario, the surface area over a specific region is calculated using a surface integral.
Surface integrals are a natural extension of double integrals, applied to represent curved surfaces in space. They help accumulate values over an area, like mass, charge, or in this case, simple surface area.
The general formula used here is:

• \[ S = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA \]
  This formula accounts for the 'tilt' of the surface in each region (which is what the derivatives help describe).

The use of this formula requires you to replace derivatives corresponding to the surface. For our example, the derivatives were constant values, making it straightforward to handle, turning the calculation particularly simple.
By setting up the integral over the triangular region, we can correctly evaluate the surface area just above the defined space.

Plane Equation

In the context of our problem, the plane equation is expressed as \(z = x + 2y\). A plane equation shows the relationship between the three coordinates - \(x\), \(y\), and \(z\) in three-dimensional space. For any given plane, the equation can take a standard form and, importantly, it reveals slope relationships:

• The coefficient of \(x\) (`1`, in this case) can be seen as a 'slope effect' along the \(x\)-axis.
• Similarly, the coefficient of \(y\) (`2`, in this example) reflects how the plane climbs along the \(y\)-axis.

Let's consider the utility of this particular form when calculating the surface area. Since the equation is z-form, it becomes directly usable in the surface integrals' formula.
Additionally, this type of linear equation tells us a lot about the overall flatness. In scenarios where the components are linear, like this example, they only involve addition, making computation and integration substantially easier than for curved surfaces.
This is a significant simplification in the study of calculus surfaces!

Partial Derivatives

Partial derivatives are incredibly valuable tools in multivariable calculus. In this exercise, we compute partial derivatives to understand how our surface behaves differently in relation to each variable \(x\) and \(y\).
When analyzing the plane \(z = x + 2y\):

• First, compute \(\frac{\partial f}{\partial x}\), which represents how \(z\) changes when \(x\) changes while \(y\) remains constant. Here, it is simply \(1\) (as \(z = x + 2y\) has a coefficient of \(x\) being \(1\)).
• Similarly, \(\frac{\partial f}{\partial y}\) is calculated, showing \(z\)'s sensitivity to changes in \(y\). This is \(2\) for our solution because doubling \(y\) doubles its influence on \(z\).

Partial derivatives are foundational in calculating other important calculus concepts, such as tangent planes and local linear approximations.
By identifying these derivative values, they were integrated into the surface area formula, providing precise understanding of surface behavior and enabling correct calculation of slope integrations. This clarity in slope contribution supports error-free calculus work throughout the process!
Partial derivatives thus not only provide snapshots of how functions change, but also underpin many integration applications.