Math 53 Worksheet Solutions - Multivariable IntegrationNote: Due to time constraints, these solutions are unusually abbreviated. Please see me inoffice hours if you would like more details.1. Find the volume of the solid in the first octant bounded by the cylinder z = 16 − x2and the plane y = 5.Solution. Region is D = [0, 4] ×[0, 5]. V =6403.2. Find the average value of f (x, y) = ey√x + eyover the rectangle R = [0, 4] × [0, 1].Solution. Area of region in question is A = 4. ThenAverage Value =14Z40Z10ey√x + eydy dx =115(4 + e)5/2− e5/2− 55/2+ 1 ≈ 3.327.3. Find the volume of the solid bounded by the planes z = x, y = x, x + y = 2, and z = 0.Solution. Discussed details at length in class. V =13.4. ComputeZZDp1 − x2− y2dAwhere D is the disk x2+ y2≤ 1, by first identifying the integral as the volume of asolid.Solution. It’s the top half of a sphere of radius one. The volume of an entire sphereof radius one is43π so,V =23π.5. Evaluate the integrals below by reversing the order of integration.(a)R√π0R√πycos(x2) dx dy(b)R10R1xex/ydy dxSolution. Draw pictures!(a)Z√π0Z√πycos(x2) dx dy =Z√π0Zx0cos(x2) dy dx = 0.1(b)Z10Z1xex/ydy dx =Z10Zy0ex/ydx dy =12(e − 1).6. Sketch the solid whose volume is given by the iterated integralZ10Z1−x20(1 − x) dy dx.Solution. Projection in xy plane is portion of parabola y = 1 −x2in the firstquadrant. Height function is z = 1 − x.7. EvaluateZZD(x2tan x + y3+ 4) dA,where D = {(x, y)|x2+ y2≤ 2}. Hint: Exploit the fact that D is symmetric withrespect to both axes.Solution. The first two terms in the integrand are odd functions of x and y,respectively. As the region is symmetric in x and y (meaning when (x, y) is in D, so is(−x, −y)), we haveZZD(x2tan x + y3+ 4) dA =ZZD4 dA = 4(π(√2)2) = 8π,where the area of the region is calculated using the formula for the area of a circle.8. Find the volume of the solid bounded by the cylinders x2+ y2= r2and y2+ z2= r2.Solution. Symmetry.V8=Zr0Z√r2−y20pr2− y2dx dy =23r3,soV
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