find the volume of the solid bounded above by the surface z=f(x,y) and below by the plane region R
f(x,y)= 4-2x-y; R{(x,y)0<x<1;0<y<2
this is pretty straightforward. Nice rectangular base.
v = ∫∫R f(x,y) dA
= ∫[0,1]∫[0,2] 4-2x-y dy dx
= ∫[0,1] (4y - 2xy - 1/2 y^2)[0,2] dx
= ∫[0,1] (6-4x) dx
= (6x - 2x^2)[0,1]
= 4
To find the volume of the solid bounded above by the surface z = f(x, y) and below by the plane region R, where f(x, y) = 4 - 2x - y, and R is the region {(x, y) 0 < x < 1; 0 < y < 2}, you can use a double integral.
The volume can be calculated using the following double integral formula:
V = ∬R f(x, y) dA
where R is the region defined by the bounds given, f(x, y) is the given surface function, and dA represents the differential area. In this case, the differential area is dx dy.
The integral can be set up as follows:
V = ∫∫R (4 - 2x - y) dx dy
To evaluate this double integral, we need to determine the limits of integration for x and y. The region R is defined as 0 < x < 1 and 0 < y < 2. Therefore, the limits of integration for the integral will be:
0 < x < 1
0 < y < 2
Now we can evaluate the double integral:
V = ∫0^1 ∫0^2 (4 - 2x - y) dx dy
First, we integrate with respect to x from 0 to 1:
V = ∫0^1 (-2x^2 - xy + 4x) |[0,2] dy
Next, we integrate with respect to y from 0 to 2:
V = ∫0^2 (-2(1)^2 - (1)y + 4(1)) - (-2(0)^2 - (0)y + 4(0)) dy
V = ∫0^2 (-2 - y + 4) dy
V = ∫0^2 (2 - y) dy
Evaluating the integral:
V = [2y - (y^2 / 2)] | [0,2]
V = (2(2) - (2^2 / 2)) - (2(0) - (0^2 / 2))
V = (4 - 2) - (0 - 0)
V = 2
Therefore, the volume of the solid bounded above by the surface z = f(x, y) and below by the plane region R is 2 cubic units.