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: