To calculate the flux across the surface SSS for the vector field F(x,y,z)\\mathbf{F}(x, y, z)F(x,y,z), we need to apply the surface integral. The vector field is given as:F(x,y,z)=\u27e8x\u2212y+z,x+y,3x+4y\u27e9\\mathbf{F}(x, y, z) = \\langle x – y + z, x + y, 3x + 4y \\rangleF(x,y,z)=\u27e8x\u2212y+z,x+y,3x+4y\u27e9<\/p>\n\n\n\n
We are asked to compute the flux \u222cSF\u22c5n\u2009dS\\iint_S \\mathbf{F} \\cdot \\mathbf{n} \\, dS\u222cS\u200bF\u22c5ndS, where n\\mathbf{n}n is the unit normal vector to the surface SSS and dSdSdS represents the surface element.<\/p>\n\n\n\n
Step 1: Surface Breakdown<\/h3>\n\n\n\n
The surface SSS consists of two parts:<\/p>\n\n\n\n
- \n
- Paraboloid<\/strong>: x2+y2+2=9x^2 + y^2 + 2 = 9×2+y2+2=9, this represents a paraboloid opening upward.<\/li>\n\n\n\n
- Disk<\/strong>: 2z=02z = 02z=0, which is a disk at z=0z = 0z=0.<\/li>\n<\/ol>\n\n\n\n
Thus, the surface consists of the upper half of a paraboloid, and the flat disk at z=0z = 0z=0.<\/p>\n\n\n\n
Step 2: Divergence Theorem<\/h3>\n\n\n\n
Since the surface SSS is closed, we can use the Divergence Theorem<\/strong> to simplify our calculations. The Divergence Theorem states:\u222cSF\u22c5n\u2009dS=\u222dV(\u2207\u22c5F)\u2009dV\\iint_S \\mathbf{F} \\cdot \\mathbf{n} \\, dS = \\iiint_V (\\nabla \\cdot \\mathbf{F}) \\, dV\u222cS\u200bF\u22c5ndS=\u222dV\u200b(\u2207\u22c5F)dV<\/p>\n\n\n\n
where VVV is the volume enclosed by SSS, and \u2207\u22c5F\\nabla \\cdot \\mathbf{F}\u2207\u22c5F is the divergence of the vector field F\\mathbf{F}F.<\/p>\n\n\n\n
Step 3: Calculate the Divergence of F\\mathbf{F}F<\/h3>\n\n\n\n
The divergence \u2207\u22c5F\\nabla \\cdot \\mathbf{F}\u2207\u22c5F is computed by taking the partial derivatives of each component of F\\mathbf{F}F:F=\u27e8x\u2212y+z,x+y,3x+4y\u27e9\\mathbf{F} = \\langle x – y + z, x + y, 3x + 4y \\rangleF=\u27e8x\u2212y+z,x+y,3x+4y\u27e9\u2207\u22c5F=\u2202\u2202x(x\u2212y+z)+\u2202\u2202y(x+y)+\u2202\u2202z(3x+4y)\\nabla \\cdot \\mathbf{F} = \\frac{\\partial}{\\partial x}(x – y + z) + \\frac{\\partial}{\\partial y}(x + y) + \\frac{\\partial}{\\partial z}(3x + 4y)\u2207\u22c5F=\u2202x\u2202\u200b(x\u2212y+z)+\u2202y\u2202\u200b(x+y)+\u2202z\u2202\u200b(3x+4y)<\/p>\n\n\n\n
Calculating the partial derivatives:\u2202\u2202x(x\u2212y+z)=1\\frac{\\partial}{\\partial x}(x – y + z) = 1\u2202x\u2202\u200b(x\u2212y+z)=1\u2202\u2202y(x+y)=1\\frac{\\partial}{\\partial y}(x + y) = 1\u2202y\u2202\u200b(x+y)=1\u2202\u2202z(3x+4y)=0\\frac{\\partial}{\\partial z}(3x + 4y) = 0\u2202z\u2202\u200b(3x+4y)=0<\/p>\n\n\n\n
Thus, the divergence is:\u2207\u22c5F=1+1+0=2\\nabla \\cdot \\mathbf{F} = 1 + 1 + 0 = 2\u2207\u22c5F=1+1+0=2<\/p>\n\n\n\n
Step 4: Set Up the Volume Integral<\/h3>\n\n\n\n
Now we can set up the volume integral. Since \u2207\u22c5F=2\\nabla \\cdot \\mathbf{F} = 2\u2207\u22c5F=2, we need to integrate over the volume VVV enclosed by the surface. The volume is the region bounded by the paraboloid and the disk at z=0z = 0z=0.<\/p>\n\n\n\n
The equation of the paraboloid is x2+y2=9\u22122zx^2 + y^2 = 9 – 2zx2+y2=9\u22122z. We will convert the region to cylindrical coordinates to simplify the integration.<\/p>\n\n\n\n
In cylindrical coordinates:<\/p>\n\n\n\n
- \n
- x=rcos\u2061\u03b8x = r \\cos \\thetax=rcos\u03b8<\/li>\n\n\n\n
- y=rsin\u2061\u03b8y = r \\sin \\thetay=rsin\u03b8<\/li>\n\n\n\n
- z=zz = zz=z<\/li>\n\n\n\n
- The volume element dV=r\u2009dz\u2009dr\u2009d\u03b8dV = r \\, dz \\, dr \\, d\\thetadV=rdzdrd\u03b8<\/li>\n<\/ul>\n\n\n\n
The bounds for the integration are:<\/p>\n\n\n\n
- \n
- rrr ranges from 0 to 3 (since x2+y2=9x^2 + y^2 = 9×2+y2=9)<\/li>\n\n\n\n
- \u03b8\\theta\u03b8 ranges from 0 to 2\u03c02\\pi2\u03c0<\/li>\n\n\n\n
- zzz ranges from 0 to 9\u2212r22\\frac{9 – r^2}{2}29\u2212r2\u200b (from the equation of the paraboloid)<\/li>\n<\/ul>\n\n\n\n
Thus, the flux is:\u222cSF\u22c5n\u2009dS=\u222dV2\u2009r\u2009dz\u2009dr\u2009d\u03b8\\iint_S \\mathbf{F} \\cdot \\mathbf{n} \\, dS = \\iiint_V 2 \\, r \\, dz \\, dr \\, d\\theta\u222cS\u200bF\u22c5ndS=\u222dV\u200b2rdzdrd\u03b8<\/p>\n\n\n\n
Step 5: Perform the Integration<\/h3>\n\n\n\n
Now, integrate over the volume VVV:\u222b02\u03c0\u222b03\u222b09\u2212r222r\u2009dz\u2009dr\u2009d\u03b8\\int_0^{2\\pi} \\int_0^3 \\int_0^{\\frac{9 – r^2}{2}} 2r \\, dz \\, dr \\, d\\theta\u222b02\u03c0\u200b\u222b03\u200b\u222b029\u2212r2\u200b\u200b2rdzdrd\u03b8<\/p>\n\n\n\n
First, integrate with respect to zzz:\u222b09\u2212r222r\u2009dz=2r(9\u2212r22)=r(9\u2212r2)\\int_0^{\\frac{9 – r^2}{2}} 2r \\, dz = 2r \\left( \\frac{9 – r^2}{2} \\right) = r(9 – r^2)\u222b029\u2212r2\u200b\u200b2rdz=2r(29\u2212r2\u200b)=r(9\u2212r2)<\/p>\n\n\n\n
Next, integrate with respect to rrr:\u222b03r(9\u2212r2)\u2009dr=\u222b03(9r\u2212r3)\u2009dr\\int_0^3 r(9 – r^2) \\, dr = \\int_0^3 (9r – r^3) \\, dr\u222b03\u200br(9\u2212r2)dr=\u222b03\u200b(9r\u2212r3)dr<\/p>\n\n\n\n
This is a standard polynomial integral:\u222b039r\u2009dr=[9r22]03=9\u00d792=40.5\\int_0^3 9r \\, dr = \\left[ \\frac{9r^2}{2} \\right]_0^3 = \\frac{9 \\times 9}{2} = 40.5\u222b03\u200b9rdr=[29r2\u200b]03\u200b=29\u00d79\u200b=40.5\u222b03r3\u2009dr=[r44]03=814=20.25\\int_0^3 r^3 \\, dr = \\left[ \\frac{r^4}{4} \\right]_0^3 = \\frac{81}{4} = 20.25\u222b03\u200br3dr=[4r4\u200b]03\u200b=481\u200b=20.25<\/p>\n\n\n\n
Thus:\u222b03(9r\u2212r3)\u2009dr=40.5\u221220.25=20.25\\int_0^3 (9r – r^3) \\, dr = 40.5 – 20.25 = 20.25\u222b03\u200b(9r\u2212r3)dr=40.5\u221220.25=20.25<\/p>\n\n\n\n
Finally, integrate with respect to \u03b8\\theta\u03b8:\u222b02\u03c020.25\u2009d\u03b8=20.25\u00d72\u03c0=40.5\u03c0\\int_0^{2\\pi} 20.25 \\, d\\theta = 20.25 \\times 2\\pi = 40.5\\pi\u222b02\u03c0\u200b20.25d\u03b8=20.25\u00d72\u03c0=40.5\u03c0<\/p>\n\n\n\n
Step 6: Final Answer<\/h3>\n\n\n\n
Thus, the flux across the surface is:40.5\u03c0\\boxed{40.5\\pi}40.5\u03c0\u200b<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"Calculate the exact flux across the surface, Jfe(x,y,z)edS for the vector field F(x,v,2)-%+li+ x-ljtzk 3x 4y where S is the closed surface consisting of two parts: Paraboloid: Disk: x^2 + y^2 + 2 = 9, x + y < 9 2z = 0 The Correct Answer and Explanation is: To calculate the flux across the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-46973","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46973","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=46973"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46973\/revisions"}],"predecessor-version":[{"id":46985,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/46973\/revisions\/46985"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=46973"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=46973"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=46973"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Disk<\/strong>: 2z=02z = 02z=0, which is a disk at z=0z = 0z=0.<\/li>\n<\/ol>\n\n\n\n