Assignment 7 - Solutions Math 209 Fall 2008

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1 Assignment 7 - Solutions Math 9 Fall 8. Sec. 5., eercise 8. Use polar coordinates to evaluate the double integral + y da, R where R is the region that lies to the left of the y-ais between the circles + y and + y. This region R can be described in polar coordinates as the set of all points r, θ with r and π/ θ 3π/. Then R + y da 3π/ π/ r cos θ + r sin θ r dr dθ 3π/ cos θ + sin θ dθ r dr π/ sin θ cos θ 3π/ r 3 π/ 3 3π/ π/ 3 r cos θ + sin θ dr dθ. Sec. 5., eercise 6. Use a double integral to find the area of the region enclosed by the curve r + 3 cos θ. This region, call it, can be described in polar coordinates as consisting of all points r, θ with r + 3 cos θ and θ π. Then Area da π +3 cos θ π r dr dθ π r π ] r+3 cos θ + 3 cos θ dθ 6 + cos θ + 9 cos θ dθ π 6 + cos θ cosθ dθ 6θ + sin θ + 9 θ + 9 ] π sinθ π. r dθ 3. Sec. 5., eercise. Use polar coordinates to find he volume of the solid inside the sphere + y + z 6 and outside the cylinder + y.

2 The sphere + y + z 6 intersects the y-plane along the circle with equation + y 6. Since the solid is symmetric about the y-plane, we may compute its total volume as twice the volume of the part that lies above the y-plane, and this latter is the solid that lies below the graph of z 6 y and above the annular region {, y + y 6}. Hence, changing polar coordinates, π Vol 6 y da 6 r r dr dθ π θ π dθ 6 r r dr 3 6 r 3/ 3 3π.. Sec. 5., part of eercise 36. Use polar coordinates to evaluate the double integral e +y da, a where a {, y + y a } is the disk of radius a centered at the origin. a e +y da, π a e r r dr dθ π a e r π e a. 5. Sec. 5.5, eercise. Electric charge is distributed over the disk + y so that the charge density at, y is σ, y + y + + y in coulombs per square meter. Find the total charge on the disk. Call Q the total charge on the disk. We have Q σ, y da + y + + y da π π π r cos θ + r sin θ + r r dr dθ π π r cos θ + sin θ + r 3 dr dθ r cos θ + sin θ dr dθ + r 3 dr dθ cos θ + sin θ dθ r dr + π r 3 dr r 3 r 3 + π 8π Coulombs. sin θ cos θ π

3 6. Sec. 5.5, eercise. Find the mass and center of mass of the lamina that occupies the region bounded by the parabolas y and y and has density function ρ, y. We may describe as a region of type I: it is the set of all points, y with y and draw a picture!. Then, calling m the mass of the lamina, we have m ρ, y da dy d d 5/ d 3. Net let s compute the first moments of the lamina: and M y dy d d 3/ 9/ d 6 55 M dy d d Finally, the center of mass of the lamina is, y My m, M m 7, / d Sec. 5.5, eercise. A lamina occupies the part of the disk + y in the first quadrant. Find its center of mass if the density at any point is proportional to the square of its distance from the origin. The region of integration part of the disk +y in the first quadrant is described easily in polar coordinates as the set of all r, θ with r and θ π. Also, the density is ρ, y k + y k + y kr, where k is a constant of proportionality. Then the mass m of the lamina is m π/ and its first moments are M y ρ, y d dy M y region region ρ, y d dy kr r dr dθ π/ π/ Finally, the center of mass is, y 8 5π, 8 5π. π/ kr 3 dr dθ kπ 8 ; kr sin θ dr dθ k 5 kr cos θ dr dθ k 5 π/ π/ sin θ dθ k 5, cos θ dθ k Sec. 5.5, eercise. Consider a square fan blade with sides of length and the lower left corner placed at the origin. If the density of the blade is ρ, y +., is it more difficult to rotate the blade about the -ais or about the y-ais?

4 We can determine in which direction rotation will be more difficult by comparing the moments of inertia about the -ais and about the y-ais. Calling the region of the y-plane occupied by the fan blade, we have I y ρ, y da y +. dy d +. d y dy +. y 3 y , 3 y and similarly I y ρ, y da y y y +. dy d +. 3 d dy Since I y > I, more force is required to rotate the blade about the y-ais; in other words, it is easier to rotate it about the -ais. 9. Sec. 5.6, eercises and 8. Evaluate the iterated integrals a y yz dz dy d b π z sin y dy dz d For a we have y while for b we have π z yz dz dy d y sin y dy dz d yz ] zy z dy d ] y y π π π π d 5 5 d 5 8 cos y ] yz y dz d cosz dz d z sinz ] z z d y 3 dy d 3 sin d π.

5 . Sec. 5.6, eercise. Evaluate the tripple integral E {, y, z, y, z }. E yz cos5 dv, where E Set this triple integral as an iterated integral: yz cos 5 dv 3 yz cos 5 dz dy d 3 y cos 5 dy d cos 5 d 3 5 sin5 ] yz cos5 3 y cos 5 3 sin. ] z z ] y y d dy d