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## Vector Calculus

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**CHAPTER 9.10~9.17**Vector Calculus**Contents**• 9.10 Double Integrals • 9.11 Double Integrals in Polar Coordinates • 9.12 Green’s Theorem • 9.13 Surface Integrals • 9.14 Stokes’ Theorem • 9.15 Triple Integrals • 9.16 Divergence Theorem • 9.17 Change of Variables in Multiple Integrals**9.10 Double Integrals**• Recall from Calculus • Region of Type ISee the region in Fig 9.71(a) R: a x b, g1(y) y g2(y) • Region of Type IISee the region in Fig 9.71(b) R: c y d, h1(x) x h2(x)**Iterated Integral**• For Type I: (4) • For Type II: (5)**THEOREM 9.12**Let f be continuous on a region R. (i) For Type I: (6)(ii) For Type II: (7) Evaluation of Double Integrals**Note:**• Volume =wherez = f(x, y) is the surface.**Example 1**Evaluate over the region bounded by y = 1, y = 2, y = x, y = −x + 5.See Fig 9.73. SolutionThe region is Type II**Example 2**Evaluate over the region in the first quadrant bounded by y = x2, x = 0, y = 4. SolutionFrom Fig 9.75(a) , it is of Type IHowever, this integral can not be computed.**Example 2 (2)**Trying Fig 9.75(b), it is of Type II**Method to Compute Center of Mass**• The coordinates of the center of mass are (10)where (11)are the moments. Besides, (x, y) is a variable density function.**Example 3**A lamina has shape as the region in the first quadrant that is bounded by the y= sin x, y = cos x between x = 0and x = 4.Find the center of mass if (x, y) = y. SolutionSee Fig 9.76.**Example 3 (5)**Hence**Moments of Inertia**• (12)are the moments of inertia about the x-axis and y-axis, respectively.**Example 4**Refer to Fig 9.77. Find Iy of the thin homogeneous disk of mass m. Fig 9.77**Example 4 (2)**Solution Since it is homogeneous, the density is the constant (x, y) = m/r2.**Radius of Gyration**• Defined by (13)In Example 4,**9.11 Double Integrals in Polar Coordinates**• Double IntegralRefer to the figure.The double integral is**Example 1**Refer to Fig 9.83. Find the center of mass wherer = 2 sin 2 in the first quadrant and is proportional to the distance from the pole. Fig 9.83**Example 1 (2)**Solution We have: 0 /2, = kr, then**Example 1 (3)**Since x = r cos andthen**Example 1 (4)**Similarly, y = r sin , then**Change of Variables**• Sometimes we would like to change the rectangular coordinates to polar coordinates for simplifying the question. If and then (3)Recall: x2 + y2 = r2 and**Example 2**Evaluate SolutionFrom the graph is shown in Fig 9.84.Using x2 + y2 = r2, then 1/(5 + x2 + y2 ) = 1/(5 + r2)**Example 2 (2)**Thus the integral becomes**Example 3**Find the volume of the solid that is under and above the region bounded byx2 + y2 – y = 0. See Fig 9.85. SolutionFig 9.85**Example 3 (2)**We find that and the equations becomeand r = sin . Now**Area**• If f(r, ) = 1,then the area is**9.12 Green’s Theorem**• Along Simply Closed Curves For different orientations for simply closed curves, please refer to Fig 9.88.Fig 9.88(a) Fig 9.88(b) Fig 9.88(c)**Notations for Integrals Along Simply Closed Curves**• We usually write them as the following formswhere and represents in the positive and negative directions, respectively.**THEOREM 9.13**• Partial ProofFor a region R is simultaneously of Type I and Type II, IF P, Q, P/y, Q/x are continuous on R, which is bounded by a simply closed curve C, then Green’s Theorem in the Plane**Partial Proof**Using Fig 9.89(a), we have**Partial Proof**Similarly, from Fig 9.89(b),From (2) + (3), we get (1).**Note:**• If the curves are more complicated such as Fig 9.90, we can still decompose R into a finite number of subregions which we can deal with. • Fig 9.90**Example 1**• Evaluate where C is shown in Fig 9.91.**Example 1 (2)**SolutionIf P(x, y) = x2 – y2, Q(x, y) = 2y – x, thenandThus**Example 2**Evaluate where C is the circle (x – 1)2 + (y – 5)2 = 4 shown in Fig 9.92.**Example 2 (2)**SolutionWe have P(x, y) = x5 + 3yandthenHence Since the area of this circle is 4, we have**Example 3**• Find the work done by F = (– 16y + sin x2)i + (4ey + 3x2)j along C shown in Fig 9.93.**Example 3 (2)**SolutionWe have Hence from Green’s theorem In view of R, it is better handled in polar coordinates, since R: