MA 351 Fall 2008 Exam #3 Review Solutions 1. (2) = λ = x 2y OR x = y = 0. = y = x 2y (2x + 2) = 2x2 + 2x 2y = 2y 2 = 2x 2 + 2x = y 2 = x 2 + x
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1 MA 5 Fall 8 Eam # Review Solutions. Find the maimum of f, y y restricted to the curve + + y. Give both the coordinates of the point and the value of f. f, y y g, y + + y f < y, > g < +, y > solve y λ + λy + + y λ y λ y OR y y y + + y y + y , y y 9 f, f, y ± f, maimum is at,. Find the dimensions of a rectangular bo of maimum volume such that the sum of the lengths of its edges is a constant, C. dimensions, y, z C + y + z z C y C maimize V yz y y V Cy y y y Cy C y when y or y C y y
2 MA 5 Fall 8 Eam # Review Solutions y is not a realistic solution for this problem y C V y C y V y C C C C + C when, C C y C z V y, V y C, V yy D >, V < a maimum C C C. Find the absolute maimum and minimum of f, y + y + y over the disk {, y + y 9} and where they occur. Interior: f + y y f y + y +, y only critical point is,, f, Boundary: f, y + y + y g, y + y 9 f < + y, + y > g <, y > solve + y λ + y λy + y 9 + y λ y λ : + λ λy λλ + λ λ λ λ 8λ + λ λ, λ,, y ± f, ± 9 λ y / y, ±
3 MA 5 Fall 8 Eam # Review Solutions f f,, λ y + y 9 ± f, f, minimum is at,, maimum is 7 at,,,. Find the volume of the solid lying under the circular paraboloid z + y and above the rectangle R,,. + y d dy + y dy 8 + y 8 y dy 6 + y dy 6 y + y Find the volume of the solid under the paraboloid z +y and above the region bounded by y and y y..5.5 The region R is of Type II:.5.5 R V y y y + y d dy + y y y y dy y + y y y + y yy dy y y 6 y 5 + y y + y y dy y y + y 5 y 6 dy y 5 y5 + y6 7 y7 5
4 MA 5 Fall 8 Eam # Review Solutions 6. Evaluate y da, D {, y y, y}. D D y da y / y dy y y dy y y d dy use u-sub on y y y / + y / dy 7. Find the volume of the solid under the paraboloid z + y + and the planes, y, z and + y. V + y + dy d y + y + y y y d + + d d 5 + d Find the volume of the solid bounded by the cylinders + y r and y + z r. By symmetry, the volume of the solid is 8 times V, which is the volume of the solid just in the first octant. The solid in the first octant is bounded by the y-plane,, y, r y and the surface z r y which in the first octant is z r y. In other words, we have V is the volume of the solid over region R pictured below under the surface z r y. r R V r r r r y r y d dy r y dy r y dy r y y r r r r r y r V 8V 6 r
5 MA 5 Fall 8 Eam # Review Solutions 5 9. Evaluate R e +y da where R {, y 6 + y 5,, y }. R e +y da { R r, θ r, 5, θ π π π 5 er e r r dr dθ r5 dθ r e 5 e 6 dθ e5 e 6 π, π }. Use polar coordinates to find the volume of the solid bounded by the paraboloid z y and the plane z. The paraboloid intersects the plane z when + y or + y r V + y da. Evaluate z y ze y d dy dz. z y +y π π π π 6π ze y d dy dz 6 r r dr dθ 6r r dr dθ r r dθ 6 π dθ dθ z ze y y dy dz z yze y dy dz ze y yz y dz ze z + z dz e z + z e e e +
6 MA 5 Fall 8 Eam # Review Solutions 6. Rewrite the integral y f, y, z dz dy d as an equivalent iterated integral in five other orders. Projection in y plane, y ^ Projection in yz plane, z y Projection in z plane,z ^ y z z y Above are the projections of the region E onto the y, yz and z-planes, respectively and are described below mathematically. D {, y,, y } {, y y,, y } D {y, z y,, z y} {y, z z,, z y } D {, z,, z } {, z z,, z } y f, y, z dz dy d E {, y, z, y D, z y, } {, y, z y, z D, y } {, y, z, z D, z y } y y y z f, y, z dz d dy y f, y, z d dz dy y f, y, z d dy dz z z z f, y, z dy dz d f, y, z dy d dz
7 MA 5 Fall 8 Eam # Review Solutions 7 π. Evaluate πyz cos 5 dv where D {, y, z, y, z }. D E D πyz cos π 5 dv πyz cos π 5 dz dy d π π yz cos 5 z dy d z π π y cos 5 π π y cos 5 dy d π π y cos 5 dy d π 5 y π y cos y d π π cos 5 d π sin 5 sin6π sin π. Evaluate dv where E is the solid enclosed by the planes z and z + y + and by the cylinders + y and + y 9. E E {r, θ, z r,, θ, π, z r cos θ + r sin θ + } dv π r cos θ+r sin θ+ π r cos θ+r sin θ+ π π π π π r cos θz r cos θr dz dr dθ r cos θ dz dr dθ zr cos θ+r sin θ+ z dr dθ r cos θ + cos θ sin θ + r cos θ dr dθ r cos θ + cos θ sin θ + r cos θ θ π r r dθ cos θ + cos θ sin θ cos θ 6 + cos θ + sin θ 65 sin θ π cos θ + 9 sin θ π cos θ dθ dθ
8 MA 5 Fall 8 Eam # Review Solutions 8 5. Evaluate E e +y +z dv, where E is enclosed by the sphere + y + z 9 in the first octant. E e +y +z dv E {ρ, θ, φ ρ,, θ, π/, φ, π/} π/ π/ π/ π/ e ρ ρ sin φ dρ dφ dθ ρ e ρ sin φ dρ dφ dθ by formula on pg 96, etended for triple integrals: int by parts twice on third integral: π/ π/ dθ sin φ dφ ρ e ρ dρ θ π/ cos φ π/ ρ e ρ ρe ρ + e ρ π 9e 6e + e + π 5e 6. Consider the two surfaces ρ csc θ in spherical coordinates and r in cylindrical coordinates. Are they the same surface, or different? Eplain your answer. Several ways to do this, this is one: For the first surface, remember: ρ cos θ sin φ y ρ sin θ sin φ z ρ cos φ For the second surface, remember: r cos θ y r sin θ z z So if we let φ in the first surface, we would get, and y. For the second surface, when, that would mean cos θ since r. Then on the second surface, by Pythagorean Identity, if cos θ, then sin θ ±. Thus y ± when. But on the first surface and y. Thus these cannot be the same surface.
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