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Key equations

  • Scalar surface integral
    S f ( x , y , z ) d S = D f ( r ( u , v ) ) | | t u × t v | | d A
  • Flux integral
    S F · N d S = S F · d S = D F ( r ( u , v ) ) · ( t u × t v ) d A

For the following exercises, determine whether the statements are true or false .

If surface S is given by { ( x , y , z ) : 0 x 1 , 0 y 1 , z = 10 } , then S f ( x , y , z ) d S = 0 1 0 1 f ( x , y , 10 ) d x d y .

True

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If surface S is given by { ( x , y , z ) : 0 x 1 , 0 y 1 , z = x } , then S f ( x , y , z ) d S = 0 1 0 1 f ( x , y , x ) d x d y .

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Surface r = v cos u , v sin u , v 2 , for 0 u π , 0 v 2 , is the same as surface r = v cos 2 u , v sin 2 u , v , for 0 u π 2 , 0 v 4 .

True

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Given the standard parameterization of a sphere, normal vectors t u × t v are outward normal vectors.

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For the following exercises, find parametric descriptions for the following surfaces.

Plane 3 x 2 y + z = 2

r ( u , v ) = u , v , 2 3 u + 2 v for u < and v < .

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Paraboloid z = x 2 + y 2 , for 0 z 9 .

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Plane 2 x 4 y + 3 z = 16

r ( u , v ) = u , v , 1 3 ( 16 2 u + 4 v ) for | u | < and | v | < .

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The frustum of cone z 2 = x 2 + y 2 , for 2 z 8

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The portion of cylinder x 2 + y 2 = 9 in the first octant, for 0 z 3

A diagram in three dimensions of a section of a cylinder with radius 3. The center of its circular top is (0,0,3). The section exists for x, y, and z between 0 and 3.

r ( u , v ) = 3 cos u , 3 sin u , v for 0 u π 2 , 0 v 3

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A cone with base radius r and height h , where r and h are positive constants

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For the following exercises, use a computer algebra system to approximate the area of the following surfaces using a parametric description of the surface.

[T] Half cylinder { ( r , θ , z ) : r = 4 , 0 θ π , 0 z 7 }

A = 87.9646

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[T] Plane z = 10 x y above square | x | 2 , | y | 2

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For the following exercises, let S be the hemisphere x 2 + y 2 + z 2 = 4 , with z 0 , and evaluate each surface integral, in the counterclockwise direction.

S z d S

S z d S = 8 π

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S ( x 2 + y 2 ) z d S

S ( x 2 + y 2 ) z d S = 16 π

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For the following exercises, evaluate S F · N d s for vector field F , where N is an outward normal vector to surface S.

F ( x , y , z ) = x i + 2 y j = 3 z k , and S is that part of plane 15 x 12 y + 3 z = 6 that lies above unit square 0 x 1 , 0 y 1 .

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F ( x , y ) = x i + y j , and S is hemisphere z = 1 x 2 y 2 .

S F · N d S = 4 π 3

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F ( x , y , z ) = x 2 i + y 2 j + z 2 k , and S is the portion of plane z = y + 1 that lies inside cylinder x 2 + y 2 = 1 .

A cylinder and an intersecting plane shown in three-dimensions. S is the portion of the plane z = y + 1 inside the cylinder x^2 + y ^2 = 1.
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For the following exercises, approximate the mass of the homogeneous lamina that has the shape of given surface S. Round to four decimal places.

[T] S is surface z = 4 x 2 y , with z 0 , x 0 , y 0 ; ξ = x .

m 13.0639

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[T] S is surface z = x 2 + y 2 , with z 1 ; ξ = z .

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[T] S is surface x 2 + y 2 + x 2 = 5 , with z 1 ; ξ = θ 2 .

m 228.5313

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Evaluate S ( y 2 z i + y 3 j + x z k ) · d S , where S is the surface of cube −1 x 1 , −1 y 1 , and 0 z 2 . in a counterclockwise direction.

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Evaluate surface integral S g d S , where g ( x , y , z ) = x z + 2 x 2 3 x y and S is the portion of plane 2 x 3 y + z = 6 that lies over unit square R : 0 x 1 , 0 y 1 .

S g d S = 3 4

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Evaluate S ( x + y + z ) d S , where S is the surface defined parametrically by R ( u , v ) = ( 2 u + v ) i + ( u 2 v ) j + ( u + 3 v ) k for 0 u 1 , and 0 v 2 .

A three-dimensional diagram of the given surface, which appears to be a steeply sloped plane stretching through the (x,y) plane.
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[T] Evaluate S ( x y 2 + z ) d S , where S is the surface defined by R ( u , v ) = u 2 i + v j + u k , 0 u 1 , 0 v 1 .

A three-dimensional diagram of the given surface, which appears to be a curve with edges parallel to the y-axis. It increases in x components and decreases in z components the further it is from the y axis.

S ( x 2 + y z ) d S 0.9617

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[T] Evaluate where S is the surface defined by R ( u , v ) = u i u 2 j + v k , 0 u 2 , 0 v 1 . for 0 u 1 , 0 v 2 .

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Evaluate S ( x 2 + y 2 ) d S , where S is the surface bounded above hemisphere z = 1 x 2 y 2 , and below by plane z = 0 .

S ( x 2 + y 2 ) d S = 4 π 3

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Evaluate S ( x 2 + y 2 + z 2 ) d S , where S is the portion of plane z = x + 1 that lies inside cylinder x 2 + y 2 = 1 .

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[T] Evaluate S x 2 z d S , where S is the portion of cone z 2 = x 2 + y 2 that lies between planes z = 1 and z = 4 .

A diagram of the given upward opening cone in three dimensions. The cone is cut by planes z=1 and z=4.

S x 2 z d S = 1023 2 π 5

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[T] Evaluate S ( x z / y ) d S , where S is the portion of cylinder x = y 2 that lies in the first octant between planes z = 0 , z = 5 , y = 1 , and y = 4 .

A diagram of the given cylinder in three-dimensions. It is cut by the planes z=0, z=5, y=1, and y=4.
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Questions & Answers

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Almas
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There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
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please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
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Professor
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Brian Reply
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industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
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LITNING Reply
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LITNING
scanning tunneling microscope
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The nanotechnology is as new science, to scale nanometric
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can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations?
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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