Use a change of variables to find the volume of the solid region lying below the surface z = f(x, y) and above the plane region R. f(x, y) = (x − y)(x + 2y) R: region bounded by the parallelogram with vertices (0, 0), (1, 1), (3, 0), (2, −1)

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 = ⅓(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 = ⅓(u - v)

Now, we have

x = ⅓(2v + u) and y = ⅓(u - v)

At this point, we calculate the partial derivatives;

dx/du, dx/dv, dy/du and dy/dv

dx/du = ⅓

dx/dv = ⅔

dy/du = ⅓

dy/dv = -⅓

Applying Jacobian;

dx/du * dy/dv - dx/dv * dy/du

= ⅓ * -⅓ - ⅔ * ⅓

= -1/9 - 2/9

= -3/9

= -⅓

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

⅔ * ⅓ * ⅔v^3/2 * 3^3/2 from 0 to 3

⅔ * ⅓ * ⅔ * (3^3/2 - 0^3/2) * 3^3/2

⅔ * ⅓ * ⅔ * 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