1-12 Find the area of the surface. The part of the surface 2 y+4 z-x^2=5 that lies above the tri

1-12 Find the area of the surface. The part of the surface 2...

Key Concepts

Surface Area of a Graph

In multivariable calculus, when a surface is described as z = f(x, y), its area above a region in the xy-plane is calculated via a surface integral. The area element is given by the formula sqrt(1 + (?f/?x)² + (?f/?y)²) dx dy. This formula adjusts the infinitesimal area in the plane to account for the slope of the surface.

Partial Derivatives

To apply the surface area integral, one must compute the first-order partial derivatives of f with respect to x and y. These derivatives measure how steeply the surface rises in the x and y directions and are squared and summed under the square root in the area formula.

Double Integration

The process of finding the total surface area requires setting up a double integral over the region in the plane. Double integration allows for the accumulation of infinitesimal area elements over the specified domain, providing a way to calculate the entire area of interest.

Integration over a Triangular Region

When the region of integration is a triangle, special attention must be given to setting up the correct limits for the double integral. This involves describing the triangle using appropriate inequalities in one of the coordinate axes, ensuring that the integration accurately covers the entire triangular domain.

Transcript

00:01 In this problem, we want to find the area of the part of the surface, which is given by 2y, plus 4 times z minus x squared, equals to 5.

00:15 That lies above the region, lies above the triangular region, with vertices 0 ,0, 2 .0, and 2 .4.

00:42 So let's start by sketching these triangular region in the x and y plane.

00:47 So here's the y axis.

00:49 Here's the x -axis.

00:51 Here's the x -axis.

00:54 And let's start by plotting the vertices that we're given to us.

01:00 So first we start off with the origin 0, then 2 .0.

01:04 So let's plot 2 .0 here.

01:07 2 .0.

01:09 Let's plot this here.

01:09 And 2 comma 4.

01:12 So 1, 2, 3, 4.

01:14 So let me put some numbers on this grid.

01:17 So here's 1, here's 2, 1, 2, 3, and 4 to make this easier.

01:22 So let's connect the vertices and we should be able to see the triangular region that they're referring to.

01:29 So here is the region that you are in there.

01:32 So if you have to describe this, the line on the top here will be y equals to 2.

01:41 And if we have to describe the limits for the variable x, x would be ranging from 0 to 2 for this region and the limits for y, y will be ranging from 0 to the line.

01:55 Y equals 2x.

01:57 We need this for the limits for the double integral.

02:01 So if we solve for z here, we can obtain our function in terms of x and y.

02:07 So let's go ahead and solve for z.

02:09 So we solve for z there, we're going to have 5 over 4 plus 1 over 4 times x squared minus 1 half times y.

02:20 And let's find partial f, partial x, partial f, partial x.

02:24 Partial f, partial x, here is just going to be 1 half times x.

02:29 And let's find partial f, partial y...