Find the average value and the surface area of z 3xx32y2y4

Find the \"average value\" and the surface area of z = 3x-x^3-2y^2+y^4 on the domain D = {(x, y)| -5<=x<= 5, -2<=y <=2}.

For average value, double integrate z w.r.t. dx and dy

integrating wrt to dx,

[1.5x^2 -0.25x^4 -2xy^2+xy^4]

putting limits of x (-5 to 5),

[0 -0 -20y^2 +10y^4] = 10 (y^4-2y^2)

integrating wrt to dy,

10(0.2y^5-(2/3)y^3)

putting limits of y (-2 to 2),

10(0.2*64 -(2/3)*(16))=21.33

So, average value= 21.33/(5+5)(2+2)=21.33/40 =0.5333

for surface area,

f_x=3-3x²=3(1-x²)

f_y=4y³-4y=4y(y²-1)

So, surface area is double integration of (1+f_x²+f_y²) with respect to dx and dy with appropriate limits

Compute this (very easy) integral with limits x= -5 to 5 and y=-2 to 2 to obtain the surface area

$Find the \$