Answer
The volume of the region lying below the surface z = f(x, y) and above the plane region R. f(x, y) = (x − y)(x + 2y) is 4
Explanation:.
f(x, y) = (x − y)(x + 2y)
vertices (0, 0), (1, 1), (3, 0), (2, −1)
Let u = x - y
v = x + 2y
We treat u like the x axis and v as the y axis. The equation from the graph are:
v = x - y = 0
u = x + 2y = 3
v = x - y = 3
u = x + 2y = 0
We need to get an equation in terms of u,v and x
Multiply v by 2 and then add to u = x + 2y
2(v = x - y)
+ u = x + 2y
----------------
2v = 2x - 2y
+ u = x + 2y
-----------------------
2v + u = 3x
Make x the subject of formula
x = â…“(2v + u)
We need to get an equation in terms of u,v and y
Subtract u = x + 2y from v = x - y
v = x - y
-u = x + 2y
---------------
v - u = -3y.
Divide through by -3
y = â…“(u - v)
Now, we have
x = â…“(2v + u) and y = â…“(u - v)
At this point, we calculate the partial derivatives;
dx/du, dx/dv, dy/du and dy/dv
dx/du = â…“
dx/dv = â…”
dy/du = â…“
dy/dv = -â…“
Applying Jacobian;
dx/du * dy/dv - dx/dv * dy/du
= â…“ * -â…“ - â…” * â…“
= -1/9 - 2/9
= -3/9
= -â…“
At the point, we have to transform the original equation using v = x - y and u = x + 2y
z = √(vu)
This is given as
∫∫z |-⅓| dudv {0,3}{0,3}
The result of the integral is the volume of the solid region
Substitute z = √(uv)
∫∫√(vu) |-⅓| dudv {0,3}{0,3}
|-⅓| ∫∫√(vu) dudv {0,3}{0,3}
⅓∫∫√v * √u dudv {0,3}{0,3} -- integrate with respect to u
⅓∫√v * ⅔u^(3/2) from 0 to 3 {0,3}
⅔ * ⅓∫√v (3^3/2 - 0^3/2) {0,3}
⅔ * ⅓∫√v . 3^3/2 {0,3}
Integrate with respect to v
â…” * â…“ * â…”v^3/2 * 3^3/2 from 0 to 3
â…” * â…“ * â…” * (3^3/2 - 0^3/2) * 3^3/2
â…” * â…“ * â…” * 3^3/2 * 3*3/2
⅔ * ⅓ * ⅔ * 3³
4
Hence, the volume of the region lying below the surface z = f(x, y) and above the plane region R. f(x, y) = (x − y)(x + 2y) is 4
