Find the surface area of the portion of a plane f(x,y)=2-x-y that lies above the circle x^2+y^2<=1 in the 1st quadrant.

To find the surface area of the portion of the plane f(x,y)=2-x-y that lies above the circle x^2+y^2<=1 in the 1st quadrant, we can use a double integral.

The first step is to draw the graph of the given function f(x,y) and the circle x^2+y^2=1 in the first quadrant. This will give us a visual representation of the problem and help us understand the limits of integration.

Next, we need to find the limits of integration for the double integral. Since we are looking for the portion of the plane above the circle x^2+y^2<=1 in the first quadrant, the limits for x and y are as follows:
- For x, the limits are from 0 to sqrt(1-y^2) (since we are in the first quadrant, x ranges from 0 to the positive x-coordinate of the point where the line y=0 intersects the circle x^2+y^2=1, which is sqrt(1-y^2)).
- For y, the limits are from 0 to 1 (since we are in the first quadrant, y ranges from 0 to 1, the positive y-coordinate of the point where the line x=0 intersects the circle x^2+y^2=1).

Now, we can set up the double integral to calculate the surface area:

Surface area = ∬R √(1 + (df/dx)^2 + (df/dy)^2) dA

Where R represents the region of integration, √ represents square root, dA represents the differential area element, and (df/dx) and (df/dy) are the partial derivatives of the function f(x,y) with respect to x and y, respectively.

In this case, the function f(x,y) is 2-x-y. Taking the partial derivatives, we get:
(df/dx) = -1
(df/dy) = -1

Substituting these into the surface area formula, we have:
Surface area = ∬R √(1 + (-1)^2 + (-1)^2) dA
= ∬R √(1 + 1 + 1) dA
= ∬R √(3) dA
= √(3) ∬R dA

Since the integrand is a constant, we can simplify the double integral to the area of the region R.

Therefore, the surface area is √(3) times the area of the region R, which is the portion of the circle x^2+y^2<=1 in the first quadrant.

To find the area of this region, we can use a single integral by integrating the top curve (the intersection of the plane and the circle) minus the bottom curve (the circle) from the lower limit (0) to the upper limit (1).

Therefore, the surface area of the portion of the plane f(x,y)=2-x-y that lies above the circle x^2+y^2<=1 in the 1st quadrant is √(3) times the integral of (2 - sqrt(1 - x^2)) - sqrt(1 - x^2) dx from 0 to 1.

To calculate the numerical value of this surface area, you can evaluate the integral using appropriate techniques such as trigonometric substitution or integral tables.