Vector Calculus Homework Check?

Question

So I am only worried about #1 right now.

So I have f(x,y,z)=2x+y+z=2,

I know P=k

So I take the gradient of F
dF/dx=2, dF/dy=1, dF/dz=1

gradient of F=2i+j+k

|gradient of F|=sqrt(4^2+1+1)=sqrt(6)

|gradient of F x P| = 1

Double Intergal of |Gradient of F| dxdy

Convert to Polar

Double Integral of |Gradient of F|rdrdtheta inner bounds 0 to 1, outter bounds 0 to 2pi

Evaluate the Integral: pi*root6

kb · Accepted Answer

1) Note that in the xy-plane, x² + y² = 2x represents a circle with radius 1/2:

x² + y² = 2x

<==> (x² - 2x) + y² = 0

<==> (x² - 2x + 1) + y² = 1

<==> (x - 1)² + y² = 1².

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So, the surface area equals

∫∫ √(1 + (∂z/∂x)² + (∂z/∂y)²) dA, using cartesian coordinates

= ∫∫ √(1 + (-2)² + (-1)²) dA, since z = 2 - 2x - y

= ∫∫ √6 dA

= √6 * (Area enclosed by x² + y² = 2x, a circle with radius 1)

= π√6.

I hope this helps!

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