Yahoo奇摩 網頁搜尋

... be the boundary of R. Consider the surface integral ∫∫_B F dot n dA, (1...n dA + ∫∫_A F dot n dA (the normal of A is downward) = Answer...

...cos(2t);;;t=0~pi/2=-3*(-1-0)=3.........ans (2)Find the area of the surface generated by revolving the curve about the x-axisAns:...

Find a tangent plane passing thru P(0,π,1) for surface :F(x,y,z)=z*sin(x)+z*cos(y)+5*z^2-7=0Sol: Normal =Grad(F)=▽F=Fn(0,π,1)=[z*cos(x);-z*sin(y);sin(x...

x= 2 p cos t y = p sin t z = exp(-4p^2) 0<=p<=1, 0<=t<= 2π is correct. ∫∫_S curl(F)∙dS = ∫_C F∙ds (By Stokes thm.) where C is the curve x^2+4y^2=4 counterclockwise...

考慮向量函數(fg, 0, 0), 由divergence theorem ∫∫_∂R (fg, 0, 0)∙n dA = ∫∫∫_R div(fg, 0, 0) dV 得 ∫∫_∂R f*g*nx dA = ∫∫∫_R ∂(fg)/∂x dV 即 ∫∫_∂R f*g*nx dA= ∫∫∫_R f(∂g/∂x)+g(∂f/∂x) dV 移項,故得證

F=(x^2, -2xy, z^2), divF=2x-2x+2z=2z ∫∫_S F‧ndA=∫∫∫_V (2z)dv =∫∫_W∫[1~2] 2zdz dxdy =∫∫_W (2^2-1^2) dxdy =3*(Area of W)=3A

Qa: F=(x^2, -2xy, z^2), div(F)=2x-2x+2z=2z ∫∫_S F‧n dA=∫∫∫_V div(F) dv (where V is the region enclosed by S) =∫∫_W∫[1~2] 2z dz da =∫∫_W (z^2)|[1~2] da =∫∫_W (2^2-1^2) da= 3∫∫_W da =3...

V is the region x^2+y^2<=z<=√(2-x^2-y^2). divF=3, by divergence thm ∫∫_S F.ndA=∫∫∫_V divF dv=3∫∫∫_V dv (choise the cylindric coord.)=3∫[0~2π]∫[0~1]∫[r^2~√(2-r^2)] dz*rdrdt =6π∫[0~1] r[√(2-r^2) - r^2] dr =6π...