MATH 253: WORKSHT 7 () Wrm-up: () Review: polr coordintes, integrls involving polr coordintes, triple Riemnn sums, triple integrls, the pplictions of triple integrls (especilly to volume), nd cylindricl coordintes. (b) Find x 2 y dx dy dz for the region [, ] [, 2] [, 3]. x 2 y dx dy dz = 3 x 2 y dz dy dx = 3x 2 y dy dx = 6x 2 dx = 2 (c) xplin why, in cylindricl coordintes (i.e., for the stndrd cylindricl coordintes with (x, y) converted to polr), r describes the distnce from the z-xis. For ech fixed constnt c, notice we re giving the horizontl slice z = c polr coordintes. In polr coordintes, r represents the distnce from point to the origin, so for point (x, y, c) in the plne z = c, r for this point represents the distnce from this point to the origin of tht plne (,, c) on the z-xis. (d) Describe the following surfces in cylindricl coordintes (gin, for stndrd cylindricl coordintes): r =, r = 4, θ =, z = 3, z = r, r = θ, z = θ. As stted bove, r represents the distnce from the z-xis, so r = represents the z-xis itself (the set of ll stuff distnce from the z-xis is just the z-xis!) nd r = 4 represents cylinder of rdius 4 wrpped round the z-xis. θ = will look like hlf of plne; notice in R 2, θ = corresponds to the x-xis, so relly we re looking t the hlf of the plne y = where x : For z = 3, notice tht when we switch into cylindricl coordintes, (x, y) chnge into polr coordintes, but z does not chnge t ll; therefore, z = 3 just looks like z = 3 in norml R 3, tht is, just horizontl plne. Dte: Mrch 2, 28.
To see wht z = r gives us, it is esiest just to chnge r bck into x nd y; this gives us z = x 2 + y 2, just our stndrd cone. For r = θ, first notice tht, since there is no dependence on z, we know tht ll the horizontl slices of this surfce will look the sme. Notice tht, s θ increses nd we move counterclockwise, r increses t the sme rte. Therefore, we get something tht spirls round the z-xis nd, since gin for ll z, we get this sme spirl, this should look like plne tht hs been twisted round the z-xis: Finlly, for z = θ, notice tht r cn be nything, so for ech choice of θ, we get line from the origin to (i.e., representing ll vlues of r) where θ = z; s θ increses, we gin rotte round counterclockwise round the z-xis while incresing z t the sme rte. This gives us the helix shpe: (2) Find the volume bounded by the xy-plne nd the prboloid z = 4 x 2 y 2 (hint: where does this surfce hit the xy-plne?). The volume under z = 4 x 2 y 2 bove the xy-plne will the volume under this grph bove the region where the grph hits the xy-plne. Notice tht 4 x 2 y 2 hits the xy-plne where z =. This is exctly where x 2 + y 2 = 4. So we wnt to find the volume bove the region D x 2 + y 2 4 below z = 4 x 2 y 2. Since D is disc, it is best to convert to polr coordintes. So our volume is given by (4 x 2 y 2 ) da = (4 r 2 )r dθdr = 2 (4r r 3 ) dr = 8 D
(3) Let be the cylinder y 2 + z 2 4 with y, z nd x. Find xz dv. Here, notice we hve cylinder wrpped round the x-xis; so, we should use version of cylindricl coordintes where we replce y nd z with r nd θ. Tking y = r cos(θ) nd z = r sin(θ), notice tht both y nd z re negtive when θ 3/2; so this is our rnge for θ. Of course, our rnge for x is from to nd our rnge for r is from to 2. So we get: xz dv = 3/2 xr sin(θ) r dx dr dθ = 3/2 r 2 sin(θ)/2 dr dθ = 3/2 4 sin(θ) dθ = 4 (4) Let be the region enclosed by the plne z = 3 nd the cone z 2 = x 2 + y 2. Find z dv. Notice tht this region is going to look like the solid cone bove z = r nd below z = 3. In cylindricl coordintes, we cn describe this either s r z (if we wnt to integrte first with respect to r) or s r z 3 (if we wnt to integrte first with respect to z). The full rnge for z would be to 3 nd the full rnge for r would lso be to 3. This gives us: 3 z 3 3 z dv = z r dr dθ dz = z 3 /2 dθ dz = z 3 dz = 8/4 (5) Clculte the volume contined by the four plnes x + y + z =, x + y + z =, y = 2, nd z = (there re few wys to do this, but you should be ble to do this with one triple integrl!). This ws probbly little tricker thn I would of liked... but here goes: I m going to use the outside in method. Notice tht one of our boundries is y = 2, so this is good indiction tht y would mke good outer bound (we could lso use z). Notice tht slice with respect to constnt vlue of y, sy y = c, must be contined by the lines x + z = c, x + z = c, nd z =, which we cn rewrite s the three lines z = x + c, z = x + c, nd z = ; so our slice is tringle formed by the three points (, c), (c, ), nd ( c, ) in the plne y = c. Since we re strting with one of our boundries s y = 2, notice this mens tht our point on the z-xis lies underneth the x-xis. So we must ssume tht z begins s negtive. To be sure we get bounded solid, we should only llow y to vry until c = (i.e., until we hit the xy-plne). This mens tht our mximum rnge for y should be y 2. Next, we need to find our inner two bounds by treting y s constnt nd nlyzing our slice for this constnt vlue of y. If we sketch the bove described tringulr slice, notice tht integrting first with respect to z would require two integrls. So we should integrte first with respect to x. this gives us rnge for x of z + y x z + y nd mximum possible rnge for z of y z.
y = So finlly, we cn clculte our volume s: z+y y z+ dx dz dy = y y 2 y + dy = 7/3 3/2 + 2y + 2z 2 dz dy = 2y(y ) ( y) 2 + (2 y) dy (6) Find the volume of the sliced cylinder contined within x 2 + y 2 4, bove the xyplne, nd below the plne z + x y = (hint: θ should not hve its full llowble rnge!) I think wht s esiest here is noticing tht you re ctully looking t the volume under the grph z = y x bove the xy-plne; so we cn ctully use double integrl! To figure out wht we should integrte over, we wnt find the projectiond onto the xy-plne. Notice tht we need x nd y to give us something in the cylinder, mening we need x 2 + y 2 4, nd we need to be under the grph, mening we need z = y x or y x. This looks like the hlf of the circle of rdius 2 sliced by y = x: 2 - -2-2 - 2 Here, we hve /4 θ 5/4 nd r 2. So we get volume of: 5/4 (y x) da = (r sin(θ) r cos(θ))r dθ dr = r 2 (2 2) dr = 6 2/3 D /4 (7) Chnge the bounds of integrtion on the triple integrl 3 y x z 2 dz dy dx to get triple integrl so tht you integrte first with respect to x, then with respect to z, nd lst with respect to y (hint: it s esiest (in my opinion) to switch one pir of djcent integrls t time). The ide here is tht the inner two bounds of integrtion correspond to the rnge for y nd z given constnt vlue of x nd the outer two bounds of integrtion correspond to the totl mximum rnge for y nd x. So, to get the required bounds
of integrtion, it is esiest to switch the inner or outer bounds one t time (this just ends up being exctly like switching the bounds of integrtion for double iterted integrls). So, to begin, let s chnge the outer two bounds. notice tht the outer two bounds describe the region x 2 nd x y 2. This gives us the tringulr region: 2..5..5...5..5 2. We cn flip the order of description with y 2 nd x y. So we cn flip our outer two bounds of integrtion: 3 y x z 2 dz dy dx = y 3 y z 2 dz dx dy To finish, we need to shift x to be the first bound of integrtion. But notice tht our inner two bounds of integrtion describe rnge for x nd z given constnt choice of y. Consequently, we cn flip these two without much work to get y 3 y z 2 dz dx dy = 3 y y z 2 dx dz dy (8) Recll tht solid of revolution is solid obtined by rotting the re under the grph of function y = f(x) round either the x or y xis. Using triple integrls (nd conversion to cylindricl coordintes), confirm the disc nd cylinder formuls: () Assuming f(x) : if the re under the curve y = f(x) between x = nd x = b is rotted round the x-xis, then the resulting solid hs volume (f(x)) 2 dx. Since we re rotting round the x-xis, we should use cylindricl coordinte system where y nd z re replced with polr coordintes. Note tht, in this cse, y = f(x) plys the role of the upper bound for the rdius of our solid; so we hve r f(x), θ 2, nd x b. Then the volume of this solid is: V = f(x) r dr dθ dx = (f(x)) 2 /2 dθ dx = f(x) 2 dx
(b) Agin, ssuming f(x) : if the re under the curve y = f(x) between x = nd x = b is rotted round the y-xis, then the resulting solid hs volume 2 xf(x) dx. Here, since we re rotting round the y-xis, we should chnge x nd z to polr coordintes. Trying to find bounds for r in terms of our other vribles would be hrd; but notice tht we cn integrte r on it s mximum rnge from to b. Our rnge for y then is y f(r) nd our rnge for θ is gin θ 2. We then get: V = f(r) rdy dθ dr = Replcing x with r yields the bove formul. rf(r) dθ dr = 2rf(r) dr