(20 points total) (a) Approximate to three decimal digits of the surface area patch on the graph ...
#P2 The differential of surface area, ds, for a surface determined by the graph of z=f(x,y) is calculated by dS = #P3 The differential of surface area, dS, for a surface determined by the graph of x = f(y,z) is calculated by dS = #P4 True or false: If Fis a velocity vector field for some fluid and S is a semipermeable surface, then the flux integral JJs FindS computes how quickly volume is passing through the surface S.
2. (20 marks) (a) Calculate the surface area of the graph of f(x,y) = x + 20y over the region R= {(x,y) e R2:1 < x < 4,2 sy s 2x} in the xy-plane. OV (b) Integrate the function g(x, y, z) = x +y +z over the surface that is described as follows: x = 2u – v, y = v + 2u, z= v – u Here u € [0,20), v € [0,21].
(14 points) Let F be the radial vector field Ft(z, y, z) =zi+w+sk And S be the surface of the cone shown at right parameterized by G(r,)-(rcos(0),r sin(0),6-3r) Write the integral F dS using an outward pointing normal in dS terms r and θ. This cone has an open bottom. . The integrand must be fully simplified » Do not evaluate the integral
(14 points) Let F be the radial vector field Ft(z, y, z) =zi+w+sk And S be the...
Directions: Use the graph to find approximate x-coordinates of the points of intersection of the given curves. Then find (approximately- three decimal places) the area of the region bounded by the curves. Also, make a rough sketch of the region sought. You must write the definite integral using proper notation to receive full credit 1) y = χ sin(x*) , y = x6
Directions: Use the graph to find approximate x-coordinates of the points of intersection of the given curves....
2. Evaluate the surface integral [[Fids. (a) F(x, y, z) - xi + yj + 2zk, S is the part of the paraboloid z - x2 + y2, 251 (b) F(x, y, z) = (z, x-z, y), S is the triangle with vertices (1,0,0), (0, 1,0), and (0,0,1), oriented downward (c) F-(y. -x,z), S is the upward helicoid parametrized by r(u, v) = (UCOS v, usin v,V), osus 2, OSVS (Hint: Tu x Ty = (sin v, -cos v, u).)...
Evaluate the surface integral lis(r,y,z) (x, y, z) ds where f(x, y, z) = x + y + z and o is the is the surface of the cube defined by the inequalities 0 < x < 5,0 Sy < 5 and 0 <3 < 5. [Hint: integrate over each face separately.] 1 f(x, y, z) ds =
(2) Let F-1 + rj + yk and consider the integral- , ▽ × F. т. dS where s is the surface of the paraboloid z = 1-12-y2 corresponding to z 0, and n is a unit normal vector to S in the positive z-direction (a) Apply Stokes' theorem to evaluate the integral. (b) Evaluate the integral directly over the surface S rectlv over the new surface
(2) Let F-1 + rj + yk and consider the integral- , ▽...
2. Evaluate 1,(1,0, 2) . ds, where s is the cone z = VE4y2 with 0 < z < 2, Upward 1,0,2) ds, where S is the pointing normal. 3. Use a surface integral to find the area of the region of the plane z2y +3 with
2. Evaluate 1,(1,0, 2) . ds, where s is the cone z = VE4y2 with 0
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2 0, y 2 0 and 0S2S1. This surface can be expressed 1 with 3. Surface S is the part of 2+y The oriented S, S, has normal vector with all three components being non-negative. Let F(, y, z) be a given vector (5 points) (1) Find the vector re xr 15 points) (2) Find the surface integral J Js ryds (3) Find the surface integral Is F ds Caution: dS but not ds (15...
Evaluate the surface integral (x2 + y' +52 ) ds where S is the part of the cone z = 2- x2 + y2 above the z = 0 plane. The surface integral equals
Evaluate the surface integral (x2 + y' +52 ) ds where S is the part of the cone z = 2- x2 + y2 above the z = 0 plane. The surface integral equals