Use the divergence theorem to evaluate s f ds where fxyz...

Use the Divergence Theorem to evaluate ∫∫S F · dS, where F(x,y,z) = (z^2)xi + ((y^3)/3+tan(z))j + ((x^2)z+(y^2))k and S is the top half of the sphere x^2 + y^2 + z^2 = 1. (Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x^2 + y^2 ≤ 1, oriented downward, and S2 is = S1 ∪ S.)

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00:01 Hello everyone, today question is let e be the semi -ball, e be the semi -ball, let we can, given the x, y, z is x squared plus y -square plus z than is equal to 1 and z greater than is equal to 0.

00:22 And we draw the rough diagram of the semi -ball, then we can evaluate the easily areas.

00:28 Then we can draw rough diagram of semibol and this find out this is area and this is s1 and this is the h2 and we can say the h2 is the h2 is the boundary of e.

00:44 We can apply for the h2 then we apply the divergence theorem.

00:50 Divergence theorem and this is the divergence theorem f dot n ds and we can get as to then we simply say that this term is equal to is equal to double integral and the volume integral of the e and divergence of f dot dv then how to find the divergence of the f then the divergence of f is given by simply this equations the divergence of f is equals to del by del x divides multiplied by z square x x plus del by del by del y by q by three plus ten j jd plus del divided by del j multiplies by x squared z plus y square terms z square plus y square plus x square plus x square then how to find we can easily we obtain the f dot n surface integral d z and is equals to volume integral of e divergence of f dot dv then puts the value of divergence of f then volume integral, divergence of x x square plus y square plus z square and dv and v simply note that the triple integral this is a volume integral can be calculated by the spherical coordinates we know that the volume integral is calculated by the it's calculated by the spherical coordinates then simply spherical coordinates the puts the values of the integrals and then makes the the indefinite to the definite integral, then we can say this term is written by 0 to 2 pi, then 0 to pi by 2 and 0 to 1...

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