Answer to Question #262984 in Calculus for sai

Question #262984

a) Let Φ(u, v) = (2 sin u cos v, 3 sin u sin v, cos u), for π/2 ≤ u, v ≤ π. Plot and describe the surface S parametrized by Φ.

b) Find the equation of the tangent plane to S at (u, v) = (3π/4, 3π/4).

c) Find the volume of the region in R3 bounded by the surface S and the coordinate planes.


1
Expert's answer
2021-11-09T16:11:01-0500

ANSWER

a)The surface S parametrized by Φ is part of the ellipsoid "\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=1" , since "\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=" "\\sin ^{ 2 }{ u } \\cos ^{ 2 }{ v+ } \\ \\sin ^{ 2 }{ u\\sin ^{ 2 }{ v+\\cos ^{ 2 }{ u\\ \\ = } \\ } }"

"=\\sin ^{ 2 }{ u } \\left( \\cos ^{ 2 }{ v+ } \\sin ^{ 2 }{ v } \\ \\right) +\\cos ^{ 2 }{ u=\\sin ^{ 2 }{ u } +\\cos ^{ 2 }{ u=1 } }"






b)The point on the surface S corresponding to (u, v) = (3π/4, 3π/4) has Cartesian coordinates "(-1,3\/2,-1\/\\sqrt { 2 } )" . Differentiating the equality "\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } +{ z }^{ 2 }=1\\quad"with respect to variables "x" and "y" we have: "{ \\left( \\frac { x }{ 2 } \\right) }+2z{ z }_{ x }^{ ' }=0\\quad ,\\quad 2{ \\left( \\frac { y }{ 9 } \\right) }+2z{ z }_{ y }^{ ' }=0" .

So, "{ z }_{ x }^{ ' }(-1,3\/2)=\\frac { \\ -1\/2 }{ 2\/\\sqrt { 2 } } =-\\ \\sqrt { 2 } \/4\\, \\quad { z }_{ y }^{ ' }(-1,3\/2)=\\frac { 1\/6 }{ \\ \\quad 1\/\\sqrt { 2 } } =\\frac { \\ \\sqrt { 2 } }{ 6 }" . Consequently, the vector "\\overrightarrow { n } =\\left< -\\frac { \\sqrt { 2 } }{ 4 } ,\\frac { \\ \\sqrt { 2 } }{ 6 } ,-1 \\right>" is a normal vector to the S at the point

"(-1,3\/2,-1\/\\sqrt { 2 } )" . Therefore, the equation of the tangent plane has the form:

"-\\frac { \\sqrt { 2 } }{ 4 } \\left( x+1 \\right) +\\frac { \\ \\sqrt { 2 } }{ 6 } \\left( y-\\frac { 3 }{ 2 } \\right) -\\left( z+\\frac { 1 }{ \\sqrt { 2 } } \\right) =0" or

"z=\\frac { -\\sqrt { 2 } x }{ 4 } +\\frac { \\ \\sqrt { 2 } y }{ 3 } -\\ \\sqrt { 2 }"

c) The region to be calculated is 1/8 of the ellipsoid . The volume of the ellipsoid "\\frac { { x }^{ 2 } }{ \\ { a }^{ 2 } } +\\frac { { y }^{ 2 } }{ { b }^{ 2 } } +\\frac { { z }^{ 2 } }{ { c }^{ 2 } } =1" is "\\frac { 4 }{ 3 } \\pi abc" .Therefore, the volume is "\\frac { 1 }{ 8 } \\cdot \\frac { 4 }{ 3 } \\pi 2\\cdot 3\\cdot 1=\\pi ."

Calculate the volume using the integral."Volume=\\ \\iiint _{ V }^{ \\ }{ dzdxdy } =\\iint _{ D }^{ \\ }{ \\left( \\sqrt { 1-\\frac { { x }^{ 2 } }{ 4 } -\\frac { { y }^{ 2 } }{ 9 } } \\right) dxdy }" , where "D=\\left\\{ (x,y):\\frac { { x }^{ 2 } }{ 4 } +\\frac { { y }^{ 2 } }{ 9 } \\le 1,\\ \\right\\} \\ \\cap \\ \\left\\{ (x,y):-2\\le x\\le 0,\\ 0\\le y\\le 3 \\right\\}". Passing to new variables "(r,v)" "x=2r\\cos { v } \\ ,\\ y=3r\\sin { v }" we get :"\\Delta =\\left\\{ (r,v):0\\le r\\le 1,\\quad \\frac { \\pi }{ 2 } \\le v\\le \\pi \\quad \\right\\}" ,

"\\iint _{ D }^{ \\ }{ \\left( \\sqrt { 1-\\frac { { x }^{ 2 } }{ 4 } -\\frac { { y }^{ 2 } }{ 9 } } \\right) dxdy } =\\iint _{ \\Delta }^{ \\ }{ \\sqrt { 1-{ r }^{ 2 } } 6r } drdv="

"=6\\int _{ \\frac { \\pi }{ 2 } }^{ \\pi }{ dv\\left( \\int _{ 0 }^{ 1 }{ \\sqrt { 1-{ r }^{ 2 } } \\ r } dr \\right) } =\\ \\frac { 3\\pi }{ 2 } \\ \\left( F(1)-F(0) \\right) =\\ \\pi" , "(F(r)=\\ { -\\frac { 2 }{ 3 } \\left( 1-{ r }^{ 2 } \\right) }^{ 3\/2 })" .


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