JfT ry dA, where D is the diamond-shaped domain (i) Use a change of variables to...

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JfT ry dA, where D is the diamond-shaped domain (i) Use a change of variables to evaluate D bounded by y = 2+1, y=1-r, y=r-1 and y=-1 – r. 6. Vector Calculus (a) Evaluate I = (r+2y+32) ds, where C = C₁+C₂. The first segment of the path, C1, is the are of the circle r² + y² = 1 lying in the ry-plane, beginning at the point (1,0,0) and ending at the point (0, 1,0). The second segment of the path, C₂, is the are of the circle y² + 2² = 1 lying in the yz-plane, beginning at the point (0, 1,0) and ending at the point (0, 0, 1). (b) Find the work done (the vector line integral [F-dr) by a particle moving through the vector field F = yi + 2²j – zk along the path C C₁ + C₂, where C₁ is the straight line segment from (1,1,1) to (2, 3, 4) and C₂ is the straight line segment from (2, 3, 4) to (0, -1,5). (e) Show that the vector field F = (e + sec r tan rtan y + y cos(ry) +2r, re³ + secr sec² y + r cos(ry) - 3y²) is con- servative, and find a potential for it. 1 1 I (d) The vector field F = (y² + + ²², 2ryz¹+· + ²¹, 2ry2¹ +1 -²², Ary²2³ - 1 + ½e²²) | is con- I Y Y Y servative (you don't need to verify it). Find a potential for F. (e) Use Green's Theorem to evaluate the vector line integral of F = (- ! + tan(e²³), In(y^) — 72² ) - over the closed curve C, which is the circle of radius 2 centered at the origin oriented in the counterclockwise direction. (f) Find the divergence and curl of F = sin(x² + y)i — e²²j + ln(z +2y+3z)k. (g) Find the flux, F-nds, of the vector field F=i-zj+rk across the surface S, where S is the portion of the plane 2+2y + z = 2 that lies in the first octant, with upward pointing normal. (h) Use Stokes' Theorem to compute ff cu curlFdS, where F = (ry, rz, yz) and S is the S part of the sphere r² + y² +2²= 9 that lies inside the cylinder r² + y² = 1 and is above the ry-plane. (i) Use the Divergence Theorem to find the net flux of F = (r, 2y, 42) out of the sphere of radius 3 centered at the origin. JfT ry dA, where D is the diamond-shaped domain (i) Use a change of variables to evaluate D bounded by y = 2+1, y=1-r, y=r-1 and y=-1 – r. 6. Vector Calculus (a) Evaluate I = (r+2y+32) ds, where C = C₁+C₂. The first segment of the path, C1, is the are of the circle r² + y² = 1 lying in the ry-plane, beginning at the point (1,0,0) and ending at the point (0, 1,0). The second segment of the path, C₂, is the are of the circle y² + 2² = 1 lying in the yz-plane, beginning at the point (0, 1,0) and ending at the point (0, 0, 1). (b) Find the work done (the vector line integral [F-dr) by a particle moving through the vector field F = yi + 2²j – zk along the path C C₁ + C₂, where C₁ is the straight line segment from (1,1,1) to (2, 3, 4) and C₂ is the straight line segment from (2, 3, 4) to (0, -1,5). (e) Show that the vector field F = (e + sec r tan rtan y + y cos(ry) +2r, re³ + secr sec² y + r cos(ry) - 3y²) is con- servative, and find a potential for it. 1 1 I (d) The vector field F = (y² + + ²², 2ryz¹+· + ²¹, 2ry2¹ +1 -²², Ary²2³ - 1 + ½e²²) | is con- I Y Y Y servative (you don't need to verify it). Find a potential for F. (e) Use Green's Theorem to evaluate the vector line integral of F = (- ! + tan(e²³), In(y^) — 72² ) - over the closed curve C, which is the circle of radius 2 centered at the origin oriented in the counterclockwise direction. (f) Find the divergence and curl of F = sin(x² + y)i — e²²j + ln(z +2y+3z)k. (g) Find the flux, F-nds, of the vector field F=i-zj+rk across the surface S, where S is the portion of the plane 2+2y + z = 2 that lies in the first octant, with upward pointing normal. (h) Use Stokes' Theorem to compute ff cu curlFdS, where F = (ry, rz, yz) and S is the S part of the sphere r² + y² +2²= 9 that lies inside the cylinder r² + y² = 1 and is above the ry-plane. (i) Use the Divergence Theorem to find the net flux of F = (r, 2y, 42) out of the sphere of radius 3 centered at the origin.

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i To evaluate ydA over the diamondshaped domain D we can use a change of variables Lets define new variables u and v such that u r y v r y We can rewrite the equations of the boundaries of D in terms ... View the full answer