[PDF] Solutions to Homework 9 integrals in cylindrical coordinates which

View - Download Solutions to Homework 9

Previous PDF

Next PDF

Solutions to Homework 9

Section 12.7 # 12: LetDbe the region bounded below by the cone z=px

2+y2and above by the paraboloidz= 2x2y2. Setup

integrals in cylindrical coordinates which compute the volume ofD.

Solution:

The intersection of the paraboloid and the cone is a circle. Since z= 2x2y2= 2r2andz=px

2+y2=r(assumingris

non-negative), 2r2=r, which implies that r

2+r2 = (r+ 2)(r1) = 0. Sincer0,r= 1. Therefore,z= 1. So,

the intersection of these surfaces is a circle of radius 1 in the planez= 1. (a) UsedV=rdz drd. The cone is the lower bound forzand the paraboloid is the upper bound forz, as is clear from a sketch of the gure. The projection (i.e. the shadow) of the region onto thexy-plane is the circle of radius 1 centered at the origin. Therefore, ZZZ D dV=Z 2 0Z 1 0Z 2r2 r rdz drd: (b) UsedV=rdrdz d. The region is not simple in ther-direction. The lower bound forris zero, but the upper bound is sometimes the conez=rand sometimes the paraboloidz= 2r2. The planez= 1 dividesDinto twor-simple regions. Therefore, ZZZ D dV=Z 2 0Z 1 0Z z 0 rdrdz d+Z 2 0Z 2 1Z p2z 0 rdrdz d: (c) UsedV=rddz dr. There is no restriction onas this region is rotationally symmetric. However,zis still constrained by the cone and the parabola. Therefore, ZZZ D dV=Z 1 0Z 2r2 rZ 2 0 rddz dr: Section 12.7 # 18: LetDbe the region enclosed by the cylindersr= cos andr= 2cosand by the planesz= 0 andz= 3y. Set up an iterated integral which computesRRR

Df(r;;z)dz rdrd.

Solution: Since 0z3y, it follows that 0z3rsinin

cylindrical coordinates. The projection ofDonto thexy-plane is the region between the circles given in polar coordinates byr= cosandr= 2cos. The rst circle is inside the second, and these two circles intersect when ==2;=2. This can be seen from a sketch or by solving the equation cos= 2cos; gather like terms to obtain 0 = cos. Therefore, ZZZ D f(r;;z)dV=Z =2 =2Z 2cos cosZ 3rsin 0 f(r;;z)rdz drd: Note: You cannot double the integral and integtrate over 0=2. Doubling is only permissible if the functionf(r;;z) is even with respect to the variable. Section 12.7 # 32:a LetDbe the region bounded below by the cone z=px

2+y2and above by the planez= 1. Set up triple integrals which

compute the volume ofD.

Solution:

(a) UsedV=2sinddd. If the pointPlies in the regionD, then varying its-coordinate keepsP insideDso long as 0sec. The upper bound is determined by the planez= 1, which has equationz=cos= 1 in spherical coordinates; solving foryields= sec. Ignoring(projecting onto= 1 for instance), one see that the variable varies from 0 to=4. Finally, since the gure is rotationlly symmetric, varies from 0 to 2. Therefore, ZZZ D dV=Z 2 0Z =4 0Z sec 0

2sinddd:

(b) UsedV=2sinddd. This integral is tricky to set up. IfPlies in the regionD, then varying its -coordinate keepsPinsideDso long as either its distance from the origin is less than or equal to one and 0=4, or its distance from the origin is greater than or equal to one and its-coordinate is bounded below by the restriction thatz= 1 and above by=4. In other words, this region is not-simple: two integrals are required. For the second integral, the conditionz= 1 implies thatcos= 1 so that sec1=4. Ignoring,, then thezcoordinate can vary from 1 top2; the upper bound is determined by the maximum distance from the origin to a point inside the regionD, which is realized by a point which lies on the intersection of the cone with the planez= 1.

The above shows that

ZZZ D dV=Z 2 0Z 1 0Z =4 0

2sinddd+Z

2 0Z p2 1Z =4 sec

1()2sinddd:

Section 12.7 # 34: Set up an integral in spherical coordinates which computes the volume of the region bounded below by the hemisphere = 1,z0, and above by the cardioid of revolution= 1 + cos. Then compute the value of the integral. Solution: Clearly 1z1 + cos. A careful sketch of the gure reveals that 0=2. This can also be determined algebraically. The cardioid and the hemisphere meet when 1 = 1 + cos, which implies that cos= 0.

Thus,ZZZ

D dV=Z 2 0Z =2 0Z 1+cos 1

2sinddd

The innermost integral evaluates to

(1 + cos)33 sinsin3 An anti-derivative with respect tocan be determined for the rst summand by using the substitutionu= 1 + cos,du=sind. The second summand is easy to integrate. Answer: 11=6. Section 12.8 # 2: Solve the systemu=x+2y,v=xyin terms ofxand yand compute the Jacobian determinant@(x;y)=@(u;v). Then sketch the image under the transformationT(x;y) = (u;v) of the triangular region in thexy-plane bounded by the linesy= 0,y=x, andx+ 2y= 2. Solution: Solving the system forxandyresults inx= (1=3)(u+ 2v) and y= (1=3)(uv). The transformed region is a triangle bounded by the line v= 0 (which corresponds tox=y), the lineu= 2 (which corresponds to x+ 2y= 2), and the lineu=v(which corresponds toy= 0). Thequotesdbs_dbs4.pdfusesText_8

[PDF] confidence interval calculator

[PDF] configuration électronique du carbone

[PDF] configuration web_url

[PDF] confinement france date du début

[PDF] confinement france voyage à l'étranger

[PDF] confinement france zone rouge

[PDF] congo two letter country code

[PDF] congressional research service f 35

[PDF] conjonctivite remède de grand mère

[PDF] conjugaison exercices pdf

[PDF] conjunction words academic writing

[PDF] connecteur de cause et conséquence

[PDF] connecteur logique juridique pdf

[PDF] connecteurs logiques dintroduction

[PDF] connecteurs logiques en francais