=(2pisqrt(2)/3) [2^(3/2) - 1] I know the first part is right but I can\'t get th

Question

=(2pisqrt(2)/3) [2^(3/2) - 1]

I know the first part is right but I can't get the second part right :/

Explanation / Answer

rs = rt = rs x rv = < -1 - 1, t+s, t-s> rs x rv = < -2, s+t, t-s> (a) at (2,3,1) s+t = 3 s-t = 1 therefore , s = 2 and t = 1 rs x rv = Therefore equation of tangent plane is -2(x-2) + 3(y-3) -1(z-1) = 0 (b) area = (double integral) |rs x rv|dsdt = (double integral) sqrt( 4 + (s+t)^2 + (s-t)^2 )dsdt = (double integral) sqrt( 2s^2 + 2t^2 + 4 )dsdt substituting s = r cosA t = r sinA for this transformation (polar transformation) dxdy = rdrdA for r^2+s^2 4rdr = du and if r = 0 to 1 then, u = 4 to 6 area = (2pi)* (integral) (1/4)sqrt(u) du from u = 4 to 6 = (2pi)*(1/4)*(2/3)( 6^3/2 - 4^3/2)

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