Let F = yi + zj + xzk. Evaluate Double integral F⋅aS for each of the following regions W: a. x2+y2 (less than or equal to) z (less thanor equal to) 1 b. x2+y2 (less than or equal to) 1 and x (greater than or equal to) 0

Question

Let F = yi + zj + xzk. Evaluate  Double integral F⋅aS for each of the following regions W:  a. x2+y2 (less than or equal to) z (less thanor equal to) 1  b. x2+y2 (less than or equal to) 1 and x (greater than or equal to) 0

Leonard Stokes · Accepted Answer

div F=x  a) The given surface x2+y2≤z≤1 is the paraboloid x2+y2=z under the plane z=1. In cylinderical co-ordinates the region bounded by the surface is given by: E=[(r,θ,z):0≤θ≤2π,0≤r≤1,r2≤z≤1]  By Gauss theorem, ∫∫SF⋅ds=∫∫∫E÷FdV  =∫02π∫01∫r21rcos⁡θr dzdrdθ  =∫02π∫01r2cos⁡θ(1−r2)drdθ  =∫02πcos⁡θdθ∫01(r2−r4)dr  [sin⁡θ]02π[r33−r55]0t=0  b) Note that here we have an extra condition x≥0 which means that we have that portion of paraboloid under the plane which lies in front ofyz-plane. Then the region boundelby this surface is:  E=[(r,θ,z]):−π2≤θ≤π2,0≤r≤1,r2≤z≤1]  ∫∫F⋅ds=∫∫∫E÷FdV  Then  =∫−π2π2∫01∫r21rcos⁡θrdzdrdθ  =[sin⁡θ]−π2π2[r33−r55]01  =2×215=415

Let F = yi + zj + xzk. Evaluate Double integral F\cdot aS for each of the following regions W: a. x^2+y^2 (less than or equal to) z (less thanor equal to) 1 b. x^2+y^2 (less than or equal to) 1 and x (greater than or equal to) 0

Answered question

Answer & Explanation

New Questions in Other