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Is the surface integral

I = âˆ« F Â· dS positive, negative, or zero for the vector field F = zi and surface S oriented as shown?

I is zero.

I is positive.

I is negative.

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07:21

Evaluate the surface integral SF · dS for the given vectorfield F and the orientedsurface S. In other words, find the fluxof F across S. For closedsurfaces, use the positive (outward) orientation.F(x, y, z)= zexyi − 3zexyj + xyk,S is the parallelogram of thisexercise with upward orientation.

04:23

Evaluate the surface integral âˆ«âˆ«âˆ« F Â· dSfor the given vectorfield F and the orientedsurface S. In other words, find the fluxof F across S. For closedsurfaces, use the positive (outward) orientation.F(x, y, z)= yi âˆ’ xj + 4zk,S is the hemisphere x^2 + y^2 + z^2 =4,z â‰¥ 0,oriented downward

07:14

Evaluate the surface integral SF · dSfor the given vector field F and theoriented surface S. In other words, find the fluxof F across S. For closedsurfaces, use the positive (outward) orientation.F(x, y, z)= xz i + x j + y kS is thehemisphere x2 + y2 + z2 = 25, y ≥ 0,oriented in the direction of the positive y-axis

06:01

Evaluate the surface integral SF · dS for the given vectorfield F and the orientedsurface S. In other words, find the fluxof F across S. For closedsurfaces, use the positive (outward) orientation.F(x, y, z)= yi − xj + 2zk, S is the hemisphere x2 + y2 + z2 =4,z ≥ 0, oriented downward

Transcript

Hi, today we are solving the question in which vector field is given by f is equals to zI which means f is equals to z 0 0. Now the surface S is oriented as shown in the figure I is e given by double integral over S f into dS. So here to determine the sign of I we need to consider the dot product f into dS. Since f is equals to z 0 and 0 the dot product will be positive. The dot product of f into dS is equals to positive when the x component of dS is positive and negative when the x component of the dS is negative and 0 when the x…

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