Double Integrals over General Regions


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1 Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing verticll: Slicing verticll mounts to slicing the intervl [, ] on the xxis, so our outer integrl will be something dx. To figure out the inner integrl, we look t generl slice. emember tht, on single slice, x is (roughl constnt, nd we wnt to describe wht does. The bottom of ech slice is on the line, nd the top is on the line x, so the inner integrl hs endpoints of integrtion nd x. Therefore, our iterted integrl is x x d dx x x ( x x x dx x dx Slicing horizontll: Slicing horizontll mounts to slicing the intervl [, ] on the xis, so our outer integrl will be something d. To figure out the inner integrl, we look t generl slice. The left end of ech slice is on the line x, nd the right end is on the line x. Since we re describing ( emember tht this is stremlined version of the rel process. ell, to get iemnn sum pproximtion, we chop the region into lots of smll rectngles, ech of width x nd height. The re of ech piece is then A x. We hve one product f(x, x per little rectngle, nd we need to dd these ll up to get iemnn sum. (See # of the worksheet Double Integrls for more detils. When converting to n iterted integrl, we re rell deciding whether we wnt to dd up in rows or columns first. If we dd up in rows, we visulize dding up in horizontl slice first nd getting one sum per horizontl slice (then we dd up ll of those sums, one per slice. Similrl, if we dd up in columns, we visulize dding up in verticl slice first nd then dding up ll those sums, one per verticl slice. So, when we s slice horizontll, we rell men we re going to dd up in rows first.
2 horizontl slice, we wnt to describe how x vries, so x goes from to. Thus, the iterted integrl is / x dx d, which is of course lso equl to.. Let be the region bounded b x nd. Write the double integrl f(x, dx d s n iterted integrl in both possible orders. Solution. Agin, we think of slicing either verticll or horizontll. x x Slicing verticll: Slicing verticll mounts to slicing the intervl [, ] on the xxis, so the outer integrl will be something dx. To write the inner integrl, we wnt to describe wht does within single slice (thinking of x s being constnt. The bottom of ech slice lies on x, nd the top lies on, so the iterted integrl is Slicing horizontll: x f(x, d dx. Slicing horizontll mounts to slicing the intervl [, ] on the xis, so the outer integrl will be something d. The left side of ech slice lies on x, nd the right side of ech slice lso lies on x. emember, though, tht we re tring to describe how x vries in slice (nd we think of s being constnt, so x goes from the left hlf of x, where x, to the right hlf, where x. Thus, the iterted integrl is f(x, dx d.. Let be the trpezoid with vertices (,, (,, (,, nd (,. Write the double integrl f(x, dx d s n iterted integrl. Solution. Let s compre slicing verticll with slicing horizontll: x x Notice tht, if we slice verticll, there re two tpes of slices. The slices to the left of x go from to, wheres the slices to the right go from to the digonl side of the trpezoid.
3 In contrst, if we slice horizontll, ll of the slices hve the sme description: the go from x to the digonl side. This seems simpler, so let s go with this method. When we slice horizontll, we re slicing the intervl [, ] on the xis, so our outer integrl will be something d. Ech slice goes from x to the digonl side. The digonl side is x (we know it s line contining the points (, nd (,. We wnt to describe how x vries in ech slice, so x goes from to. So, the iterted integrl is. Evlute the double integrl f(x, dx d. ( + dx d where is the region in the first qudrnt bounded b x,, nd x. (To decide the order of integrtion, first think bout whether it s esier to integrte the integrnd with respect to x or with respect to. Solution. The integrnd is much esier to integrte with respect to x thn with respect to. Therefore, we should tr to rewrite the double integrl s n iterted integrl where the inner integrl is with respect to x. This mens our outer integrl will be with respect to, which corresponds in our strteg to slicing the region horizontll. x This mounts to slicing the intervl [, ] on the xis, so the outer integrl will be something d. Ech slice hs its left end on x nd its right end on x. We wnt to describe how x vries within slice, so we rewrite x s x. This gives the iterted integrl + dx d ( x + + d x x We cn evlute this integrl using substitution: if we let u +, then du d, nd we cn rewrite the integrl s u u du 9 u/ u ( / 9 d ( If ou used the other order of integrtion, ou should hve sum of iterted integrls x f(x, d dx. f(x, d dx +
4 5. In ech prt, ou re given n iterted integrl. Sketch the region of integrtion, nd then chnge the order of integrtion. ( x f(x, d dx. Solution. Let s just think of our strteg in reverse. The fct tht the outer integrl is something dx tells us tht we re slicing the intervl [, ] on the xxis, so we re mking verticl slices from x to x. The inner integrl tells us tht the bottom of ech slice is on, nd the top of ech slice is on x. So, the region of integrtion (with verticl slices looks like the picture on the left: x x x x (b To chnge the order of integrtion, we wnt to insted use horizontl slices (the picture on the right. Now, we re slicing the intervl [, ] on the xis, so the outer integrl is something d. Ech slice hs its left edge on x (or x, since we rell wnt to describe x in terms of nd its right edge on x, so we cn rewrite the iterted integrl s f(x, dx d. Solution. The fct tht the outer integrl is f(x, dx d. something d tells us tht we re slicing the intervl [, ] on the xis, so we re mking horizontl slices from to. The inner integrl tells us tht the left side of ech slice is on x nd the right side is on x (or x. So, the region of integrtion looks like this: x x x x To chnge the order of integrtion, we use verticl slices. Now, we re slicing the intervl [, ] on the xxis. The bottom of ech slice is on x, nd the top of ech slice is on, so we cn rewrite the integrl s x f(x, d dx.
5 (c f(x, dx d. Solution. The fct tht the outer integrl is something d tells us tht we re slicing the intervl [, ] on the xis, so we re mking horizontl slices from to. The inner integrl tells us tht the left side of ech slice is on x nd the right side of ech slice is on x. x describes the left hlf of the circle x +, nd x describes the right hlf, so the region of integrtion looks like this: x x To chnge the order of integrtion, we use verticl slices. Now, we re slicing the intervl [, ] on the xxis, so the outer integrl is something dx. Ech slice hs its bottom edge on nd its top edge on the top hlf of the circle x + (or x, so we cn rewrite the iterted integrl s x f(x, d dx. 6. Let be constnt between nd. Let be the region bounded b x + nd. Write the double integrl f(x, dx d s n iterted integrl in both possible orders. Solution. The curves x + nd intersect where x, so x ±. So, the region looks like this:,, x To write the double integrl s n iterted integrl, we think of slicing either verticll or horizontll. 5
6 ,,,, x x Slicing verticll: Slicing verticll corresponds to slicing the intervl [, ] on the xxis, so the outer integrl will be edge on, so the iterted integrl is something dx. Ech slice hs its bottom edge on x + nd its top constnt, so it s fine to hve it in the outer integrl. Slicing horizontll: x + f(x, d dx. emember tht is Slicing horizontll corresponds to slicing the intervl [, ] on the xis, so the outer integrl will be something d. Ech slice hs its left edge on x + (so x nd its right edge on x + (so x. Thus, the iterted integrl is 7. Evlute the iterted integrl x x cos ( d dx. f(x, dx d. Solution. We don t know how to integrte the integrnd with respect to, but we cn integrte it with respect to x. This suggests tht we should chnge the order of integrtion, s in??. First, let s figure out wht the region looks like. The fct tht the outer integrl is something dx tells us tht we re slicing the intervl [, ] on the xxis, so we re mking verticl slices from x to x. The inner integrl tells us tht the bottom of ech slice is on x (the bottom hlf of the circle x + nd the top of ech slice is on. So, the region of integrtion looks like this: x x 6
7 To chnge the order of integrtion, we switch to using horizontl slices. Now, we re slicing the intervl [, ] on the xis, so our outer integrl will be something d. Ech slice hs its left edge on x nd its right edge on the right hlf of the circle x + (so x. Therefore, we cn rewrite the given integrl s x cos ( dx d x cos ( d ( cos ( We cn use substitution to evlute this integrl: let u ; then, du ( d, so the integrl becomes cos u du sin u / sin u u / ( d 8. A flt plte is in the shpe of the region in the first qudrnt bounded b x,, ln x nd. If the densit of the plte t point (x, is xe grms per cm, find the mss of the plte. (Suppose the x nd xes re mrked in cm. Solution. As we lerned in #(b of the worksheet Double Integrls, we cn find the mss of the plte b tking the double integrl of the densit, where the region of integrtion is the plte. In this cse, the integrnd xe is es to integrte with respect to x nd with respect to, so we will pick n order of integrtion bsed on the shpe of the region. We cn either slice horizontll or verticll: x x As in??, this region is simpler to describe using horizontl slices: with verticl slices, there re two tpes of slices, but with horizontl slices, there is onl one. If we use horizontl slices, we re slicing the intervl [, ] on the xis. Ech slice goes from x to ln x (or x e, so the iterted integrl is e ( xe xe dx d x e d e d 6 e 6 ( e 6 x 7
8 9. Let U be the solid bove z, below z, nd between the surfces x sin nd x sin +. Find the volume of U. Solution. The picture on the left shows the four surfces z, z, x sin, nd x sin +. The picture on the right shows just the solid U. z z x x This solid cn be described s the solid under z over the region, where is where the solid meets the xplne. So, its volume will just be ( dx d. To clculte this double integrl, we need to describe nd convert the double integrl to n iterted integrl. The surfce z intersects the xplne z where, or ±, so nd re boundx dries of the region. The other two re x sin nd x sin +. So, looks like this: x sin x x sin It s esier to slice this region horizontll: x sin x x sin This mounts to slicing the intervl [, ] on the xis, so the outer integrl will be something d. 8
9 The left side of ech slice is on x sin, nd the right side is on x sin +, so we cn rewrite the double integrl s n iterted integrl sin + sin ( dx d 8 [ x( ( d xsin + xsin ] d 6 9
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