The area of the surface is 144.708
The equation of the given surface is,
z=g(x,y)=xy
Solving the partial derivatives,
∂g∂x=y,∂g∂y=x
Substituting to the formula
S=∬√1+( ∂g∂x)2+(∂g∂y)2dA
Thus,
S=∬√1+(y)2+(x)2dAS=∬√1+x2+y2dA
The region in the XY-plane is defined by the intervals 0≤θ≤2π,0≤r≤4
Converting the integral into polar coordinates,
S=∫2π0∫40√1+r2rdrdθ
Integrating with respect to r
S=∫2π0[13(1+r2)32]40dθ
S=∫2π0(17√173−13)dθ
Integrating with respect to θ
S=(17√173−13)[θ]2π0
S≈144.708
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3 pounds = 0.003 killograms
Answer:
![[11.5-(2.5*3)]^2=16](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D16)
Step-by-step explanation:
We want to evaluate ![[11.5-(2.5*3)]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2)
Let us evaluate within the parenthesis first:
![[11.5-(2.5*3)]^2=[11.5-(7.5)]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B11.5-%287.5%29%5D%5E2)
![\implies [11.5-(2.5*3)]^2=[11.5-7.5]^2](https://tex.z-dn.net/?f=%5Cimplies%20%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B11.5-7.5%5D%5E2)
We again subtract within the bracket to obtain:
![[11.5-(2.5*3)]^2=[4]^2](https://tex.z-dn.net/?f=%5B11.5-%282.5%2A3%29%5D%5E2%3D%5B4%5D%5E2)
This finally gives us:
Answer:
4.5
Step-by-step explanation:
