334. Find the value of divergence theorem for the field D = 2xy i + x2 j for the rectangular parallelepiped given by x = 0 and 1, y = 0 and 2, z = 0 and 3.
Ratna yadavEnlightened
Check your spam folder if password reset mail not showing in inbox????
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
b Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
b
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx
dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
B Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12
B
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx
dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12
Answer: b Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Answer: b
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx
dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Answer: b Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
Answer: b
See lessExplanation: While evaluating surface integral, there has to be two variables and one
constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx
dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
b) Explanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.
b)
See lessExplanation: While evaluating surface integral, there has to be two variables and one constant compulsorily. ∫∫D.ds = ∫∫Dx=0 dy dz + ∫∫Dx=1 dy dz + ∫∫Dy=0 dx dz + ∫∫Dy=2 dx dz + ∫∫Dz=0 dy dx + ∫∫Dz=3 dy dx. Put D in equation, the integral value we get is 12.