![]() ![]() In other words, the surface is given by a vector-valued function P (encoding the x, y, and z coordinates of points on the surface) depending on two parameters, say u and v. Recall that a surface is an object in 3-dimensional space Downloaded by Fares Amzarih ( ) lOMoARcPSD|23476781 Uploaded by ChiefHedgehog3111 on coursehero.In this lesson, we will study integrals over parametrized surfaces. We have z x = y and z y = x, so dS = p y 2 + x 2 +1. Solution: The surface S sits above the region D in the xy -plane defined in polar coordinates by 1 ≤ r ≤ 2, 0 ≤ t ≤ 2 π. Let S be the portion of the surface z = xy that is inside the cylinder x 2 + y 2 =4 but outside the cylinder x 2 + y 2 =1. We parametrize this circle by x ( t ) = √ 3cos( - t ), y ( t ) = √ 3sin( - t ), z ( t ) = 3 for 0 ≤ t ≤ 2 π. ![]() The boundary of S is the upper rim, which is the circle C defined by x 2 + y 2 = 3 and z = 3, oriented clockwise so that the outer side of S is on the left as we move around the circle. We observe that if G ( x,y,z )= - y i - z j - 2 x k then curl( G )=(0+1) i +(2+0) j +(0+1) k = F so we may apply Stokes' Theorem. The surface S 2 sits above the region D in the xy -plane defined in polar form by 1 ≤ r ≤ √ 3, 0 ≤ θ ≤ 2 π. Downloaded by Fares Amzarih ( ) lOMoARcPSD|23476781įor the side S 2, we use the outward normal n =. ![]() n dS = ZZ S 1 - dS = - (area of S 1 )= - π.Since S is oriented outward, we use the normal n = and obtain ZZ S 1 F The bottom of the bowl is the disc S 1 defined by z =1 and x 2 + y 2 ≤ 1. The top of the bowl is open, and S is oriented outwards. Let S be the bowl having a flat base in the plane z = 1 and whose sides are the part of the paraboloid z = x 2 + y 2 between z =1 and z =3. ![]() Parametrize the line C 3 from (2, 2) to (1, 1) by x =2 - t, y =2 - t, 0 ≤ t ≤ 1. Parametrize the line C 2 from (3, 1) to (2, 2) by x =3 - t, y =1+ t, 0 ≤ t ≤ 1. Parametrize the line C 1 from (1, 1) to (3, 1) by x = t, y = 1, 1 ≤ t ≤ 3. d r = ZZ T (2 x - 2) dA = Z 2 1 Z 4 - y y (2 x - 2) dxdy = Z 2 1 ( x 2 - 2 x ) vextendsingle vextendsingle 4 - y y dy = Z 2 1 ((4 - y ) 2 - 2(4 - y ) - y 2 +2 y ) dy = Z 2 1 (8 - 4 y ) dy = (8 y - 2 y 2 ) vextendsingle vextendsingle 2 1 =16 - 8 - 8+2= 2.Theleftsideof T istheline y = x andtherightsideistheline x + y =4. If C is the closed path around T in the counter-clockwise direction and F =2 y i + x 2 j, evaluate the line integral Z C F Let T be the triangle with vertices at (1, 1), (3, 1), and (2, 2). Downloaded by Fares Amzarih ( ) lOMoARcPSD|23476781ģ. I = Z 1 0 Z 2 x x p 1+ x 2 dydx = Z 1 0 (2 x p 1+ x 2 - x p 1+ x 2 ) dx = Z 1 0 x p 1+ x 2 dx = 1 3 (1+ x 2 ) 3 / 2 vextendsingle vextendsingle vextendsingle vextendsingle 1 0 = 1 3 (2 √ 2 - 1). 0.2 0.4 0.6 0.8 1.0 0.5 1.0 1.5 2.0 We change the order of integration to dydx and evaluate. The dashed line is y =2 x, the solid line is y = x. Solution: The region of integration is shown below. Evaluate Z 1 0 Z y y/ 2 p 1+ x 2 dxdy + Z 2 1 Z 1 y/ 2 p 1+ x 2 dxdy. Then the mass is, using polar coordinates, ZZ D ρ ( x,y ) dA = Z 2 π 0 Z 1 0 (2+ r 2 ) r drd θ = Z 2 π 0 ✓ r 2 + r 4 4 ◆vextendsingle vextendsingle vextendsingle vextendsingle 1 0 =2 π ✓ 1+ 1 4 ◆ = 5 π 2 g. Solution: Let D be the region x 2 + y 2 ≤ 1. Suppose a lamina L (flat plate) occupies the region defined by x 2 + y 2 ≤ 1 and has density ρ ( x,y )=2+ x 2 + y 2, with ρ in g/cm 2 and x, y in cm. A formula sheet appears on the back of this page. Only ENCS approved calculators are permitted. Answers should be given as exact values (e.g. Macdonald Time: 90 minutes MaAll solutions must include a carefully written explanation. ENGR 233-S Midterm 2 Concordia University Instructor: J. ![]()
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