Consider the surfaces $S _ { 1 }$ : $\frac { 1 } { 9 } x ^ { 2 } + \frac { 1 } { 9 } y ^ { 2 } + \frac { 1 } { 4 } z ^ { 2 } s = 1 , z \geq 0$ , and $S _ { 2 }$ : $4 z = 9 - x ^ { 2 } - y ^ { 2 } , \quad z \geq 0$ , and let F be a vector field with continuous partial derivatives everywhere. Why do we know that $\iint _ { S _ { 1 } } \operatorname { curl } \mathbf { F } \cdot d \mathbf { S } = \iint _ { S _ { 2 } } \operatorname { curl } \mathbf { F } \cdot d \mathbf { S }$ ?

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The Answer of Consider the surfaces $S _ { 1 }$ : $\frac { 1 } { 9 } x ^ { 2 } + \frac { 1 } { 9 } y ^ { 2 } + \frac { 1 } { 4 } z ^ { 2 } s = 1 , z \geq 0$ , and $S _ { 2 }$ : $4 z = 9 - x ^ { 2 } - y ^ { 2 } , \quad z \geq 0$ , and let F be a vector field with continuous partial derivatives everywhere. Why do we know that $\iint _ { S _ { 1 } } \operatorname { curl } \mathbf { F } \cdot d \mathbf { S } = \iint _ { S _ { 2 } } \operatorname { curl } \mathbf { F } \cdot d \mathbf { S }$ ?

Consider the Surfaces S1S _ { 1 }S1​ 19x2+19y2+14z2s=1,z≥0\frac { 1 } { 9 } x ^ { 2 } + \frac { 1 } { 9 } y ^ { 2 } + \frac { 1 } { 4 } z ^ { 2 } s = 1 , z \geq 091​x2+91​y2+41​z2s=1,z≥0

Consider the Surfaces $S _ { 1 }$ $\frac { 1 } { 9 } x ^ { 2 } + \frac { 1 } { 9 } y ^ { 2 } + \frac { 1 } { 4 } z ^ { 2 } s = 1 , z \geq 0$