Question

Find the area of the surface. the part of the plane 2x 3y z = 6 that lies in the first octant

Answer 1

To calculate the area of the surface of the plane 2x + 3y + z = 6 in the first octant, establish a double integral using y and z as variables and integrate over the appropriate bounds.

Step-by-step explanation:

To find the area of the portion of the plane 2x + 3y + z = 6 that lies in the first octant, we need to set up the appropriate integral. The first octant implies that x, y, z are all non-negative. The plane intercepts the x-axis when y=0 and z=0, the y-axis when x=0 and z=0, and the z-axis when x=0 and y=0. Solving these intercepts, we find:

• For the x-intercept: (3,0,0)
• For the y-intercept: (0,2,0)
• For the z-intercept: (0,0,6)

Now, we can establish a double integral to find the area. We choose y and z as our variables of integration, and solve for x from the plane equation, yielding x = (6 - 3y - z)/2. The limits of integration for y are from 0 to 2 (from the y-intercept) and for z from 0 to 6 - 3y (from the z-intercept equation when x is zero).

The integral to find the area A is:

A = ∫ ∫ x dy dz

Substitute x and put the integral bounds:

A = ∫_0^2 ∫_0^{6-3y} (6 - 3y - z)/2 dz dy

Work out the inner integral first (with respect to z), then the outer integral (with respect to y), and you will obtain the area of the surface in the first octant.

Answer 2

To find the area of the surface that lies in the first octant of the plane 2x + 3y + z = 6, you can determine the boundaries of the first octant and calculate the area using a rectangular region.

Step-by-step explanation:

To find the area of the surface that lies in the first octant of the plane 2x + 3y + z = 6, we can first determine the boundaries of the first octant. The first octant is defined by positive values of x, y, and z.

To find these boundaries, we set x, y, and z equal to zero in the equation 2x + 3y + z = 6, one at a time. By doing so, we find that the boundary values for the first octant are x = 0, y = 0, and z = 0.

Next, we substitute the values of these boundaries into the equation to find the corresponding values of the other variables. Plugging in x = 0, y = 0, and z = 0, we get 0 + 0 + z = 6, which simplifies to z = 6.

Therefore, the area of the part of the plane that lies in the first octant is the region between the coordinates (0, 0, 0) and (0, 0, 6). Since this is a rectangular region, we can find the area by multiplying the difference in the z-coordinates by the product of the x and y distances, which is 0.