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 first quadrant, we need to calculate the area of the region between the plane and the circle.

First, let's visualize the problem. The plane f(x, y) = 2 - x - y intersects the circle x^2 + y^2 = 1 in the first quadrant. This intersection forms a circular disk-like shape.

To find the surface area, we'll consider the circular disk as a collection of tiny infinitesimal surface areas. We'll integrate the area element dA over the region.

Now, let's find the intersection point(s) of the plane and the circle.

Setting the equation of the circle, x^2 + y^2 = 1, equal to the equation of the plane, 2 - x - y = 0, we can solve this as a system of equations to find the intersection point(s).

Solving the system:
2 - x - y = 0
x^2 + y^2 = 1

Let's substitute x = 2 - y into the second equation:

(2 - y)^2 + y^2 = 1
4 - 4y + y^2 + y^2 = 1
2y^2 - 4y + 3 = 0

Solving this quadratic equation, we obtain two solutions: y = 1 and y = 1/2.

Using these values of y, we can determine the corresponding x-values from the equation of the plane:

For y = 1, we have:
2 - x - 1 = 0
x = 1

For y = 1/2, we have:
2 - x - 1/2 = 0
x = 3/2

So, we have two intersection points: (1, 1) and (3/2, 1/2).

To calculate the surface area between the plane and the circle in the first quadrant, we'll use the formula for the surface area of a portion of a surface of revolution:

A = ∫[a, b] 2πf(x)√(1 + (f'(x))^2) dx

Since we're working in the first quadrant, the limits of integration will be x = 1 to x = 3/2.

Let's find f'(x), the derivative of f(x, y):

f(x, y) = 2 - x - y
∂f/∂x = -1

The derivative doesn't depend on x or y, so ∂f/∂x = -1.

Now, let's substitute the values of x, y, and f'(x) into the formula to calculate the surface area:

A = ∫[1, 3/2] 2π(2 - x - y)√(1 + (-1)^2) dx
A = ∫[1, 3/2] 2π(2 - x - y)√2 dx

Since the equation is symmetric about the y-axis, the surface area in the first quadrant will be one-fourth of the total surface area. Therefore, we need to multiply the result by 4 to get the final answer.

At this point, you can either solve the integral analytically or numerically using a calculator or software to find the surface area.

Once you calculate the value of the integral, multiply it by 4 to get the final surface area of the portion of the plane above the circle in the first quadrant.