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Suppose W Consists of the Interior of Two Intersecting Cylinders x2+z2=4,y=5x ^ { 2 } + z ^ { 2 } = 4 , y = 5

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Suppose W consists of the interior of two intersecting cylinders of radius 2.One cylinder is centered on the y-axis and extends from y = -5 to y = 5.The other is centered on the x-axis and extends from x = -5 to x = 5.Let S be the entire surface of W except for one circular end of one cylinder, namely the circular end centered at (0,5,0).The boundary of S is therefore the circle x2+z2=4,y=5x ^ { 2 } + z ^ { 2 } = 4 , y = 5 ; the surface S is oriented outward.
Let F=(3x2+3z2)j=curl(z3i+y3jx3k)\vec { F } = \left( 3 x ^ { 2 } + 3 z ^ { 2 } \right) \vec { j } = \operatorname { curl } \left( z ^ { 3 } \vec { i } + y ^ { 3 } \vec { j } - x ^ { 3 } \vec { k } \right) .
Then SFdA=Qπ\int _ { S } \vec { F } \cdot d \vec { A } = Q \pi .Find Q.


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