MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.3. Double Integrals 3A. Double Integrals in Rectangular Coordinates 3A-1 Evaluate each of the following iterated integrals: (6x2 +2y) dy dx b) lnI2 Ln(, sin t +t cos U) dt du 3A-2 Express each double integral over the given region R as an iterated integral, using the given order of integration. Use the method described in Notes I to supply the limits of integration. For some of them, it may be necessary to break the integral up into two parts. In each case, begin by sketching the region. a) R is the triangle with vertices at the origin, (0,2), and (-2,2). Express as an iterated integral: i) dy dx ii) /L dx dy b) R is the finite region between the /L parabola y = 2x -/Lx2 and the x-axis. Express as an iterated integral: i) dy dx ii) dx dy c) R is the sector of the circle with center /L at the origin and radius 2 lying between the x-axis and the line u " =x.Express as an iterated integral: i) dy dx ii) /L dx dy d)* R is the finite region lying between the parabola y2 = x and the line through (2,O) having slope 1. Express as an iterated integral: i) /L dy dx ii) dx dy 3A-3 Evaluate each of the following double integrals over the indicated region R. Choose whichever order of integration seems easier -this will be influenced both by the integrand, and by the shape of R. a) /L x dA; R is the finite region bounded by the axes and 2y +x = 2 b) /L(2x +Y2) dA; R is the finite region in the first quadrant bounded by the axes and y2 = 1-x; (dx dy is easier). c) /L y dA; R is the triangle with vertices at (f1, O), (0,l). 3A-4 Find by double integration the volume of the following solids. a) the solid lying under the graph of z = sin2 x and over the region R bounded below by the x-axis and above by the central arch of the graph of cosx b) the solid lying over the finite region R in the first quadrant between the graphs of x and x2, and underneath the graph of z =xy.. c) the finite solid lying underneath the graph of x2 -y2, above the xy-plane, and between the planes x = 0 and x = 12 E. 18.02 EXERCISES 3A-5 Evaluate each of the following iterated integrals, by changing the order of integration (begin Jd2by l2figuring out what the region 1;' R is, and sketching it). a) e-Y2dy dx b) Jdli4 $dudt dudx 3A-6 Each integral below is over the disc consisting of the interior R of the unit circle, centered at the origin. For each integral, use the symmetries of R and the integrand i) to identify its value as zero; or if its value is not zero, ii) to find a double integral which is equivalent (i.e., has the same value), but which has a simpler integrand and/or is taken over the first quadrant (if possible), or over a half-disc. (Do not evaluate the integral.) 3A-7 By using the inequality f 5 g on R + J& f dA 5 JJR gdA, show the following estimates JJ,are valid: d A < a) +area of R 1+JJ,~4 ~4 -x dA < .35, R is the square 0 <+ +x, y 1.b, i x2 y2 <3B. Double Integrals in Polar Coordinates In evaluating the integrals, the following definite integrals will be useful: 1.3.5.....(n-l)n-if n is an even integer 2 22.4. .. .. . n 2'sinnx dx = cosnx dx = 2.4 .....(n -1) 1.3 ........n ' if n is an odd integer 2 3. For example: Jd and the same holds if cosx is substituted for sinx. 3B-1 Express each double integral over the given region R as an iterated integral in polar coordinates. Use the method described in Notes I to supply the limits of integration. For some of them, it may be necessary to break the integral up into two parts. In each case, begin by sketching the region. a) The region lying to the right of the circle with center at the origin and radius 2. and to the left of the vertical line through (-1,O). b)* The circle of radius 1, and center at (0,l). c) The region inside the cardioid T = 1-cos8 and outside the circle of radius 312 and center at the origin. d) The finite region bounded by the y-axis, the line y =a, and a quarter of the circle of radius a and center at (a, 0).3. DOUBLE INTEGRALS 3 3B-2 Evaluate by iteration the double integrals over the indicated regions. Use polar coordinates. /LF; R is the region inside the first-quadrant loop of r = sin20.a) R is the first-quadrant portion of the interior of x2 + SLy2 = a2 C) tan2 0 dA; R is the triangle with vertices at (0, O), (1, O), (1,l). SL dx dy -R is the right half-disk of radius $ centered at (0,;). 3B-3 Find the volumes of the following domains by integrating in polar coordinates: a) a solid hemisphere of radius a (place it so its base lies over the circle x2 + y2 = a2) b) the domain under the graph of xy and over the quarter-disc region R of 3B-2b c) the domain lying under the cone z = d m and over the circle of radius one and center at (0,l) d) the domain lying under the paraboloid z = x2 + y2 and over the interior of the right-hand loop of r2 = cos 0. 3B-4* Sometimes students wonder if you can do a double integral in polar coordinates iterating in the opposite order: J& d0 dr . Though this is uncommon, just to see if you can carry out in a new situation the basic procedure for putting in the limits, try supplying the limits for this integral over the region bounded above by the lines x = 1 and y = 1, and below by a quarter of the circle of radius 1and center at the origin. 3C. Applications of Double Integration If no coordinate system is specified for use, you can use either rectangular or polar coordinates, whichever is easier. In some of the problems, a good placement of the figure in the coordinate system simplifies the integration a lot. 3C-1 Let R be a right triangle, with legs both of length a, and density 1. Find the following ((b) and (c) can be deduced from (a) with no further calculation) a) its moment of inertia about a leg; b) its polar moment of inertia about the right-angle vertex; c) its moment of inertia about the hypotenuse. 3C-2 Find the center of mass of the region inside one arch of sinx, if: a) 6 = 1 b) 6 = y 3C-3 D is a diameter of a disc of radius a, and C is a chord parallel to D with distance c from it. C divides the disc into two segments; let R …
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