Evaluating a Flux Integral In Exercises 31 and 32, find the flux of F over the closed surface. (Let N be the outward unit normal
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Chapter 15 Solutions
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- U (x, y) 4c(y + 1)i + xyj, and V(x, y) = Consider two vector fields in the xy plane, given in the Cartesian coordinates as: cyi - xj, where c is a constant. Find where in the xy plane the vectors of these two fields are parallel to one another, and where they are mutually orthogonal. =arrow_forwardUsing Green's Theorem, find the outward flux of F across the closed curve c. F= (x? + y2)i+ (x-y)j; C is the rectangle with vertices at (0,0), (4,0), (4,8), and (0,8) А. 96 В. -224 C. 288 D. 160arrow_forwardEvaluate Curlvñ , where v= 2xyi +(x² – 2y)j+ xzk and ñ is a unit vector normal to the surface shown in the figure: (i) (ii) surface z=i surface y=1arrow_forward
- The figure in this exercise shows a horizontal layer of the vector field of a fluid flow in which the flow is parallel to the xy-plane at every point and is identical in each layer (i.e., is independent of z). State whether you believe that the curl is nonzero at the origin. If you believe that it is nonzero, then state whether it points in the positive or negative z-direction. X O The curl at the origin is zero. O The curl at the origin is nonzero and points in the negative z-direction. O The curl at the origin is nonzero and points in the positive z-direction.arrow_forwardEXERCISE 5 The temperature is T degrees at any point (x, y, z) in three-dimensional space and T(x,y, z) = 1/(x² + y² + z² + 3). %3D Distance is measured in inches. (a) Find the rate of change of the temperature at the point (3, –2, 2) in the direction of the vector –2 i+3j-6k. (6) Find the direction and magnitude of the greatest rate of change of T at (3,–2,2). 14arrow_forwardwww Find the circulation of F = -yi + x²j + zk around the oriented boundary of the part of the paraboloid z = 9 - x² - y² lying above the xy-plane and having the normal vector pointing upward.arrow_forward
- Find a unit normal vector to the surface at the given point. [Hint: Normalize the gradient vector VF(x, y, z).] Surface Point z = V x2 + (8, 6, 10) Part 1 of 6 The equation of the surface can be converted to the general form by defining F(x, y, z) as F(x, y, z) = Vx2 + y2 – E The gradient of F is the vector given by VF(x, y, z) = Fx(x, y, z)i + F (x, y, z)j + F (х, у, z)k. Part 2 of 6 Determine the partial derivatives Fx(x, y, z), Fy(x, y, z), and F2(x, y, z). Filx, y, z) = (V + y? - z) x2 FAx, Y, ») = V+yå - 2) ду x² + y2 Fo(X, Y, 2) = (V + y? - 2) azarrow_forwardTutorial Exercise Find a unit normal vector to the surface at the given point. [Hint: Normalize the gradient vector VF(x, y, z).] Surface Point x² + y2 (8, 6, 10) z = Part 1 of 6 The equation of the surface can be converted to the general form by defining F(x, y, z) as F(x, y, z) = v x2 + y2 The gradient of F is the vector given by VF(x, y, z) = Fx(x, y, z)i + F (х, у, 2)j + F (х, у, 2)k.arrow_forwardSubject differential geometry Let X(u,v)=(vcosu,vsinu,u) be the coordinate patch of a surface of M. A) find a normal and tangent vector field of M on patch X B) q=(1,0,1) is the point on this patch?why? C) find the tangent plane of the TpM at the point p=(0,0,0) of Marrow_forward
- Compute the flux of the vector field F(x, y, z) = 3i +2j+ 2k through the rectangular region with corners at (1, 0, 0), (1, 1, 0), (1, 1, 2), and (1, 0, 2) oriented in the positive x- direction, as shown in the figure. Flux =arrow_forwardFind a unit normal vector to the surface at the given point. [Hint: Normalize the gradient vector VF(x, y, z).] Surface Point (8, 6, 10) Part 1 of 6 The equation of the surface can be converted to the general form by defining F(x, y, z) as F(x, Y, z) = Vx + y? The gradient of F is the vector given by VF(x, y, z) = F,(x, y, z)i + F (х, у, 2)j + F (х, у, 2)k. Part 2 of 6 Determine the partial derivatives F(x, y, z), F(x, y, z), and F-(x, y, z). F(x, y, z) = V? + y? Fy(x, y, z) = F-(x, y, z) = vx² + y² - Part 3 of 6 Substitute the partial derivatives in the gradient and evaluate at the given point. VF(x, y, z) - i-k VF(8, 6, 10) = 4/5 4/5 3/5 j - k Part 4 of 6 If F is differentiable at (x, y, z) and VF(x, y, z) + 0, then VF(x, y, z) is normal normal to the level curve through (x, y, z). Hence, to obtain a unit normal vector n to the surface, normalize the gradient vector. Normalization of a vector is the process of obtaining a unit vector parallel parallel to the direction of the original…arrow_forwardFind r(t) · u(t). r(t) = (5t – 3)i + t³j + 2k u(t) = t2i – 6j + t3k r(t) · u(t) = Is the result a vector-valued function? Explain. Yes, the dot product is a vector-valued function. No, the dot product is a scalar-valued function.arrow_forward
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