91Ó°ÊÓ

When dealing with functions of two variables, such as the surface defined by \( z = 1 + 3x + 2y^2 \), taking partial derivatives allows us to understand how the function changes as one variable changes, while the other stays constant. A partial derivative gives the rate at which a function changes with respect to one variable, if we keep all other variables constant.

Think of it as slicing the surface in the direction of the variable you are differentiating about. In this exercise:

• The partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), is calculated as 3. This means that for every unit increase in \( x \), \( z \) increases by 3, holding \( y \) constant.
• The partial derivative with respect to \( y \), denoted \( \frac{\partial f}{\partial y} \), is calculated as \( 4y \). This indicates that the increase or decrease in \( z \) depends linearly on \( y \), so as \( y \) changes, the slope of \( z \) will vary accordingly.

In this problem, to find the surface area of the portion of the plane, we use a double integral. A double integral allows us to sum up small elements over a two-dimensional area. It is much like adding up slices or infinitesimally small rectangles on the surface.

When you establish the double integral for surface area, such as:\[\int_0^2 \int_0^{\frac{x}{2}} \sqrt{10 + 16y^2} \space dy \space dx\]you are indicating that you want to add up elements of surface area from \( y = 0 \) to \( y = \frac{x}{2} \), for each \( x \) from 0 to 2. This process aggregates to get the total area above the triangle in the xy-plane converted by the function into 3D space.

Understanding which variable to integrate first (\( dy \) or \( dx \)) can depend on the region and the function. Here, you integrate \( y \) first because its limits are expressed in terms of \( x \).

Determining the bounds of integration is crucial for correctly evaluating double integrals over a specified region, especially when dealing with surfaces. In high-dimensional calculus, bounds specify where the integrals start and end, defining the shape or area over which you are integrating.

In this exercise, the region of integration is the triangle defined by the vertices \((0,0)\), \((0,1)\), and \((2,1)\). This region lies in the xy-plane and describes the foundation of the 3D shape whose surface area is to be calculated.

• For \( x \), the bounds are from 0 to 2, matching the boundaries of the triangle along the x-axis.
• For \( y \), because of the triangle's slope, the lines within the region alters as \( x \) changes, leading \( y \) to vary from 0 to \( \frac{x}{2} \).

These bounds specify the area in xy-space over which the surface, defined by the function, projects.

The concept of a surface area integral is to calculate the area of a surface in 3D space. Such calculations are necessary in situations involving physical surfaces such as the exterior of complex shapes or graphical representations in engineering and science.

In the formula for the differential element of a surface area:\[ dS = \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \space dx \space dy\]the expression under the square root accounts for the slope of the surface in both directions, effectively compensating for how much more 'surface' is present due to tilting away from the xy-plane.

In this exercise, the surface is \( z = 1 + 3x + 2y^2 \). When calculating the integral, after determining partial derivatives, you found:\[\sqrt{10 + 16y^2}\]This factor is integrated over your defined region to compute the whole surface area, capturing the actual geometry of the surface.