EECS+145+HW3

# EECS+145+HW3 - M TOPICS IN INTEGRATION 569 x2 dA = r4...

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Unformatted text preview: M TOPICS IN INTEGRATION 569 x2}: dA = r4 ceszﬂ sin 9 d9 dr r=ﬂ ﬂ=ﬂ Furthermore, since the limits ef integratien with respect te 9 de net invelve r, we can write -1 2n Ely dA : r4 (11' ceslﬂ sin 9 d9 r=ﬂ E=ﬂ R and the denble integral in this case reduces te a precinct ef integrals. Thus we ebtain fy (1/1 = [; r5]; Hassle]? = e : WW“ a“ tr- ?” Triﬂe-s “a! a?" a t "If rﬁt ,. W?“ “is its -" a "ff-t serif . a i " if??? is“? F sites a .c .. i t ﬁﬁgﬂhﬁaﬁ a e as h a I-ifagtﬁt' «'3 " F '35??? 3??” if ' “3531’ w' as as Its” trﬁttatssaameﬁatv ﬁt we “start t“ ’ r a t r- ..,»..;-ra afresh?! at? waist aw Evaluate the fellewing: 4 Evaluate If sin :1th + y) dx dy ever the triangle whese vertices are (0, 0), (2, l), (l, 2). 3 a - s s 2 (3) “Wm +y) dy d‘x (b) x y dy d‘x Sketch the dernains ef integratien ef the denble e . I 2 1 integrals l 2 l 1 2 2 (c) (2): +3.» )dy dx (3) dx ﬂ 4 _..1 -2 I} x MIII(1 ) 1H? P Evaluate . 2 . 2 (b) dy (ces 2y)a(1 — I: am a)dx 1 0 U x— dx (1 y y Change the erder ef integratien, and hence evaluate the integrals. ever the rectangle bennded by the lines .r = 0, Wan x = 2, y = 1 and y = 2_ Evaluate 1 l a 2- ~ . dx Evaluate If (a: + y ) ex rly ever the regien fer which dy 1 xE0,y;0andx+yE1. e r'rtElyU'E-xn Sketch the dernain ef integratien ef the deuble integral Sketch the demain ef integratien and evaluate 2 2.1: 1 1...: . 2 (a) d]: Edy 2 dx (x +y) l gilII—x } x 1 x'x -%J: H ﬂ , g g dJHjx I; D H J(xI-FJF) 1 a I— l (e) div: 1 dy Express the integral in pelar ceerdinates, and hence a 2 a can?) mm '“x ‘J’ ) shew that its value is 574 VECTOR CALCULUS Figure 7.28 Three-dimensional generalization of Green’s theorem. -' . 3787. ' Surface 5' where dS = r2 dS is the vector element of surface area and ii is a unit vector along the normal. This generalization is called Stokes‘ theorem, and will be discussed in Section 7.4.12 after we have formally introduced the concept of a surface integral. Evaluate the line integral i¥[sinydx+(x—cosy)dy] C taken in the anticlockwise sense, where C is the perimeter of the triangle formed by the lines l l y=§rcx, y=im x=0 Verify your answer using Green’s theorem in a plane. Use Green’s theorem in a piano to evaluate it [of —y) dx+ (Hy?) dy] C as a double integral, where C is the triangle with vertices at (0, 0), (2, O) and (2, 2) and is traversed in the anticlockwise direction. Evaluate the line integral I = % (xydx+xdy) C where C is the closed curve consisting ofy = x2 fromx=0tox=1andy= V‘xfromx= l tox=0. 7.4.7 Surface integrals Conﬁrm your answer by applying Green‘s theorem in the plane and evaluating I as a double integral. Use Green’s theorem in a plane to evaluate the line integral final 3;?) dx + (ey+ 4x2) dy] C where C is the circle x2 +-y2 : 4. (Hint: use polar coordinates to evaluate the double integral.) Evaluate a 2a—x dx % d}, 0 1 4a + { y + x)' using the transformation of coordinates u = x + y, v=x—y Using the transformation x + y = u, — = 12 show that 1 2—y 4- 2 l dej x 2y6x+ydx=J clqu e“dv=e2—l 0 y x o 0 _________.___d.__________________________________m_._______________...___________._____H.u____________m..______________.____..._________________ The extensions of the idea of an integral to line and double integrals are not the only generalizations that can be made. We can also extend the idea to integration over a general surface S. Two types of such integrals occur: M Topics IN INTEGRATION 531 Evaluate the area of the surface .s = 2 — x2 — y: lying Evaluate the surface integral ﬁg U(x, y, 2:) d3, 11. above the (x, y) plane. (Hint: Use polar coordinates I?” i where S is the surface of the paraboloid - I to evaluate the double integral.) a = 2 — (xii + ya) above the (x, y) plane and U(x, y, s) is given by - Evaluate (a) ffﬂxi + yi)dS, where S is the surface area of . . . . . the plane 23; + y + 23 = 6 cut me by the planes Give a physical interpretation in each case. e=ﬂ,z=2,y=0,y=3; (1)) HS: d8, where Sis the surface area of the hemisphere x2 + y2 + a2 = 1 (2 1‘s- 0) cut off by the cylinder 3:2 — x + y2 = 0. (a) 1 (lirr).:ri+r2 (c): Determine the surface area of the plane 2x+y+2z=16cutoffbyx=0,y=0 andx2+y2= 64. Show that the area of that portion of the surface of the paraboloid x2 + y2 = 4: included between (a) ti: (xy, 11:2, .1: + s) andS is the part of the planes: =1 ands = 3 is ‘13? 1:04 — 3'2}. the plane 2x + 2y + s = 6 included in the ﬁrst octant; Evaluate the surface integral in Example 7.26 using 03) v = (3y, 2x3, .33) and S is the surface of the cylindrical polar coordinates. cylinders:2 +y2 =1, 0 1:: a < 1. Evaluate I f 31: ~ d3, where If F = yr? + (x — 23:21,? — xyk, evaluate the surface Show that ﬂgsi d5 = in, where Sis the surface of integral ff3(cur1F)- d5, where S is the surface of the sphere is2 +y2 + a2 =1, e E U. the sphere is: +y2 + a2 = erg, s w 0. 7.4.9 Volume integrals In Section 7.4.? we deﬁned the integral of a function over a curved surface in three dimensions. This idea can be extended to deﬁne the integral of a function of three variables through a region T of three-dimensional Space by the limit . _ . ' _ - ' I _ . - . ' - . ' - a _ - . -_ -- -- -. ._“--':da_-.-'_--r-.n--- -.--c--.-.,-.' .H'.--'-.'_a':":'.- -'.r-_£-':'eh ae'ni-hi'l'iﬁ'haﬁﬁ- of -_-'.-_- :14— .-.o.. where AV]. (E = 1, . . . , n) is a partition of T into n elements of volume, and (iii-\$1.51.) is a point in AK. as illustrated in Figure 7.36. In terms of rectangular cartesian coordinates the triple integral can, as illustrated in Figure 7.37, be written as b gzix} hgixiy} f (x: y... 2) fl V = dx dy for, y, z) (12' (7.32) :1 Elite} alter] 1" Note that there are six different orders in which the integration in (7.32) can be carried out. .5;- As we saw for double integrals in (7.28), the expression for the element of volume " dV = dx dy dz under the transformation a: = x(u, v, w), y = y(u, u. w), z' = z(u, I}, w) may be obtained using the Jacobian 584 VECTOR CALCULUS ________________._.._—...—.-——--—-—-~——‘-— In this example it is natural to use spherical polar coordinates, so that r = 3M3 (r2 sin29) r2 sin9 dr d6 c1¢ 41! a sphere H II = i”; r4dr sin36 d6 41m 0 0 2 =§Ma W Evaluate the triple integrals i 2 3 (MJ de dyJ x2};de 0 0 I 2 3 4 (mi j [ xyzzdzdydx (l l 2 Show that I z .H-Z J, £12]de (x+y+z)dy=0 7] (l x—z Evaluate III sin (x + y + z) (1.: dy dz over the portion of the positive octant cut off by the plane x + y + z 2 1:. Evaluate LUV xyz dx d y dz, where V is the region bounded by the planes )5 = 0, y = 0, z = 0 and x + y + z = 1. Sketch the region contained between the parabolic cylinders y z x2 and x 2 y2 and the planes 2 : 0 and x + y + z = 2. Show that the volume of the region may be expressed as the triple integral 1 :1 Z—x—y J J J dz dy dx 0 x2 0 and evaluate it. Use spherical polar coordinates to evaluate “’eroc2 + y2 + 22) dx dydz V 2 0 (up = 3M,(§a5)<§)(2n) 411a where Vis the region in the ﬁrst octant lying within the sphere x2 +322 +22 = 1. Evaluate Hf xzyzziix + y + 2) dx dy dz thr0ughout the region deﬁned byx+y+z \$1,): a 0,y 2 0, z B 0. Showthatifx+y+z=u,y+2=uuaridz=uvw then Hence evaluate the triple integral exp[~4(x + y + 2):] dx dy dz V where Vis the volume of the tetrahedron bounded bytheplanesx=0,y=0,z=0andx+y+z= 1. Evaluate HIV yz dx dydz taken throughout the prism with sides parallel to the z axis, whose base is the triangle with vertices at (0, 0, O), (1, 0, 0), (0, 1, 0) and whose top is the triangle with vertices at (0, 0, 2), (1, O, 1), (0, 1, 1). Find also the position of the centroid of this prism. Evaluate HI 2 dx dy dz thmughout the region deﬁnedbyx2+y2 S 22,x2+yz+z2 S1,z > 0. Using spherical polar coordinates, evaluate III x dx dy d2 throughout the positive octant of the sphere x2 +y2 + 22 = a2. M ...
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EECS+145+HW3 - M TOPICS IN INTEGRATION 569 x2 dA = r4...

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