How to evaluate the integrals in the cylindrical coordinates

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Evaluate the following integral in cylindrical coordinates
$$int^1_-1int^sqrt1-x^2_0int^2_0dfrac11+x^2+y^2dzdydx$$




My try:



I first took the boundaries as $$-1le xle1\0le ylesqrt1-x^2\0le zle2$$
and I know the formula that $$D=(r,theta,z):g(theta)le rle h(theta),alphalethetalebeta,G(x,y)le zle H(x,y)$$$$int^_int^_Dint^_f(r,theta,z)dV=int^beta_alphaint^h(theta)_g(theta)int^H(rcosalpha,rsintheta)_G(rcostheta,rsintheta)f(r,theta,z)dzdrdtheta$$



But how to apply this formula and change the boundaries of the integrals?



Can anyone please explain this.






calculus integration multivariable-calculus definite-integrals cylindrical-coordinates







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    up vote
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    Evaluate the following integral in cylindrical coordinates
    $$int^1_-1int^sqrt1-x^2_0int^2_0dfrac11+x^2+y^2dzdydx$$




    My try:



    I first took the boundaries as $$-1le xle1\0le ylesqrt1-x^2\0le zle2$$
    and I know the formula that $$D=(r,theta,z):g(theta)le rle h(theta),alphalethetalebeta,G(x,y)le zle H(x,y)$$$$int^_int^_Dint^_f(r,theta,z)dV=int^beta_alphaint^h(theta)_g(theta)int^H(rcosalpha,rsintheta)_G(rcostheta,rsintheta)f(r,theta,z)dzdrdtheta$$



    But how to apply this formula and change the boundaries of the integrals?



    Can anyone please explain this.






    calculus integration multivariable-calculus definite-integrals cylindrical-coordinates







    share|cite|improve this question







    New contributor




    user982787 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite












      Evaluate the following integral in cylindrical coordinates
      $$int^1_-1int^sqrt1-x^2_0int^2_0dfrac11+x^2+y^2dzdydx$$




      My try:



      I first took the boundaries as $$-1le xle1\0le ylesqrt1-x^2\0le zle2$$
      and I know the formula that $$D=(r,theta,z):g(theta)le rle h(theta),alphalethetalebeta,G(x,y)le zle H(x,y)$$$$int^_int^_Dint^_f(r,theta,z)dV=int^beta_alphaint^h(theta)_g(theta)int^H(rcosalpha,rsintheta)_G(rcostheta,rsintheta)f(r,theta,z)dzdrdtheta$$



      But how to apply this formula and change the boundaries of the integrals?



      Can anyone please explain this.






      calculus integration multivariable-calculus definite-integrals cylindrical-coordinates







      share|cite|improve this question







      New contributor




      user982787 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.












      Evaluate the following integral in cylindrical coordinates
      $$int^1_-1int^sqrt1-x^2_0int^2_0dfrac11+x^2+y^2dzdydx$$




      My try:



      I first took the boundaries as $$-1le xle1\0le ylesqrt1-x^2\0le zle2$$
      and I know the formula that $$D=(r,theta,z):g(theta)le rle h(theta),alphalethetalebeta,G(x,y)le zle H(x,y)$$$$int^_int^_Dint^_f(r,theta,z)dV=int^beta_alphaint^h(theta)_g(theta)int^H(rcosalpha,rsintheta)_G(rcostheta,rsintheta)f(r,theta,z)dzdrdtheta$$



      But how to apply this formula and change the boundaries of the integrals?



      Can anyone please explain this.






      calculus integration multivariable-calculus definite-integrals cylindrical-coordinates




      calculus integration multivariable-calculus definite-integrals cylindrical-coordinates






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          3 Answers
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          First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0le thetale pi$ and $0le rle 1$.



          So:



          $$int_0^piint_0^1int_0^2frac11+r^2rdzdrdtheta$$



          This integral should be easy enough, use a $u$ sub for the $r$ part.






          share|cite|improve this answer








          New contributor




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          • Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
            – user982787
            4 hours ago

















          up vote
          2
          down vote













          I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.



          Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$iiint_D f(r,theta,z)rdzdrdtheta$$



          The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $theta$, the $r$ bounds will depend on $theta$, and the $theta$ bounds will be constants.



          In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.






          share|cite|improve this answer




















          • Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
            – user982787
            3 hours ago

















          up vote
          0
          down vote













          Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$int^2_0 left( int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydxright)dz$$
          Let $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$



          Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):



          enter image description here



          Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case).



          We substitute $x=rcostheta$ and $y=rsintheta$ so that $x^2+y^2=r^2$



          Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $dtheta$ in angular direction. Thus the length of small differential element is $rdtheta$ and therefore differential area element is $rdrdtheta$.That is, $$dA = dxdy = rdrdtheta$$
          enter image description here



          We can see that the limits for $r$ and $theta$ for the semicircle will be,



          $$0le rle1\0le thetalepi$$



          Therefore,
          $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$
          $$I=int^1_0int^pi_0dfrac11+r^2rdrdtheta$$






          share|cite|improve this answer






















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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

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            active

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            up vote
            2
            down vote













            First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0le thetale pi$ and $0le rle 1$.



            So:



            $$int_0^piint_0^1int_0^2frac11+r^2rdzdrdtheta$$



            This integral should be easy enough, use a $u$ sub for the $r$ part.






            share|cite|improve this answer








            New contributor




            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.

















            • Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
              – user982787
              4 hours ago














            up vote
            2
            down vote













            First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0le thetale pi$ and $0le rle 1$.



            So:



            $$int_0^piint_0^1int_0^2frac11+r^2rdzdrdtheta$$



            This integral should be easy enough, use a $u$ sub for the $r$ part.






            share|cite|improve this answer








            New contributor




            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.

















            • Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
              – user982787
              4 hours ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0le thetale pi$ and $0le rle 1$.



            So:



            $$int_0^piint_0^1int_0^2frac11+r^2rdzdrdtheta$$



            This integral should be easy enough, use a $u$ sub for the $r$ part.






            share|cite|improve this answer








            New contributor




            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            First of all, your $z$ goes from $0$ to $2$ so we just have to look at the $xy$ plane. In the $xy$ plane, you have the upper half circle with radius $1$. So $0le thetale pi$ and $0le rle 1$.



            So:



            $$int_0^piint_0^1int_0^2frac11+r^2rdzdrdtheta$$



            This integral should be easy enough, use a $u$ sub for the $r$ part.







            share|cite|improve this answer








            New contributor




            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 5 hours ago









            Zachary Selk

            178110




            178110




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            New contributor





            Zachary Selk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            • Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
              – user982787
              4 hours ago
















            • Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
              – user982787
              4 hours ago















            Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
            – user982787
            4 hours ago




            Can you explain how you got the intervals $0lethetalepi$ and $0le rle1$ with the help of diagram if possible.
            – user982787
            4 hours ago










            up vote
            2
            down vote













            I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.



            Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$iiint_D f(r,theta,z)rdzdrdtheta$$



            The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $theta$, the $r$ bounds will depend on $theta$, and the $theta$ bounds will be constants.



            In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.






            share|cite|improve this answer




















            • Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
              – user982787
              3 hours ago














            up vote
            2
            down vote













            I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.



            Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$iiint_D f(r,theta,z)rdzdrdtheta$$



            The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $theta$, the $r$ bounds will depend on $theta$, and the $theta$ bounds will be constants.



            In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.






            share|cite|improve this answer




















            • Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
              – user982787
              3 hours ago












            up vote
            2
            down vote










            up vote
            2
            down vote









            I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.



            Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$iiint_D f(r,theta,z)rdzdrdtheta$$



            The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $theta$, the $r$ bounds will depend on $theta$, and the $theta$ bounds will be constants.



            In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.






            share|cite|improve this answer












            I think your primary confusion here arises from the fact that you think you must "apply" that formula. What I believe that formula is telling you is the general form for a cylindrical variable transformation.



            Side note: I believe the argument of the integral requires an additional $r$ in it; i.e. it should say $$iiint_D f(r,theta,z)rdzdrdtheta$$



            The main takeaway from your formula should be that after your transformation, the $z$ bounds will depend on $r$ and $theta$, the $r$ bounds will depend on $theta$, and the $theta$ bounds will be constants.



            In order to determine the specific bounds for your problem, you should try to picture what the domain $D$ of this integral looks like. Here, it is just a half-cylinder of height 2, so all of your bounds will be constants, as Zachary has already pointed out.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            BSplitter

            443215




            443215











            • Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
              – user982787
              3 hours ago
















            • Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
              – user982787
              3 hours ago















            Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
            – user982787
            3 hours ago




            Can you explain in Zachary's answer about how he got the intervals $0≤θ≤π$ and $0≤r≤1$ with the help of diagram if possible.
            – user982787
            3 hours ago










            up vote
            0
            down vote













            Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$int^2_0 left( int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydxright)dz$$
            Let $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$



            Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):



            enter image description here



            Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case).



            We substitute $x=rcostheta$ and $y=rsintheta$ so that $x^2+y^2=r^2$



            Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $dtheta$ in angular direction. Thus the length of small differential element is $rdtheta$ and therefore differential area element is $rdrdtheta$.That is, $$dA = dxdy = rdrdtheta$$
            enter image description here



            We can see that the limits for $r$ and $theta$ for the semicircle will be,



            $$0le rle1\0le thetalepi$$



            Therefore,
            $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$
            $$I=int^1_0int^pi_0dfrac11+r^2rdrdtheta$$






            share|cite|improve this answer


























              up vote
              0
              down vote













              Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$int^2_0 left( int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydxright)dz$$
              Let $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$



              Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):



              enter image description here



              Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case).



              We substitute $x=rcostheta$ and $y=rsintheta$ so that $x^2+y^2=r^2$



              Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $dtheta$ in angular direction. Thus the length of small differential element is $rdtheta$ and therefore differential area element is $rdrdtheta$.That is, $$dA = dxdy = rdrdtheta$$
              enter image description here



              We can see that the limits for $r$ and $theta$ for the semicircle will be,



              $$0le rle1\0le thetalepi$$



              Therefore,
              $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$
              $$I=int^1_0int^pi_0dfrac11+r^2rdrdtheta$$






              share|cite|improve this answer
























                up vote
                0
                down vote










                up vote
                0
                down vote









                Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$int^2_0 left( int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydxright)dz$$
                Let $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$



                Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):



                enter image description here



                Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case).



                We substitute $x=rcostheta$ and $y=rsintheta$ so that $x^2+y^2=r^2$



                Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $dtheta$ in angular direction. Thus the length of small differential element is $rdtheta$ and therefore differential area element is $rdrdtheta$.That is, $$dA = dxdy = rdrdtheta$$
                enter image description here



                We can see that the limits for $r$ and $theta$ for the semicircle will be,



                $$0le rle1\0le thetalepi$$



                Therefore,
                $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$
                $$I=int^1_0int^pi_0dfrac11+r^2rdrdtheta$$






                share|cite|improve this answer














                Since we see that the limits of $z$ are independent of $x$ and $y$, we can rewrite the given integral as:$$int^2_0 left( int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydxright)dz$$
                Let $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$



                Now, observing the limits of $x$ and $y$ it can be seen that they span over the area of a semicircle (as shown in the image below):



                enter image description here



                Therefore, to evaluate $I$ we'll convert the integral into polar cooordinates (which are cylindrical coordinates in this case).



                We substitute $x=rcostheta$ and $y=rsintheta$ so that $x^2+y^2=r^2$



                Now we have to find $dA=dxdy$ in these coordinates. In polar coordinates let us consider a strip of width $dr$ in radial direction $dtheta$ in angular direction. Thus the length of small differential element is $rdtheta$ and therefore differential area element is $rdrdtheta$.That is, $$dA = dxdy = rdrdtheta$$
                enter image description here



                We can see that the limits for $r$ and $theta$ for the semicircle will be,



                $$0le rle1\0le thetalepi$$



                Therefore,
                $$I=int^1_-1int^sqrt1-x^2_0dfrac11+x^2+y^2dydx$$
                $$I=int^1_0int^pi_0dfrac11+r^2rdrdtheta$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                yathish

                217111




                217111




















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                    What does second last employer means? [closed]

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                    Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
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                    List of Gilmore Girls characters

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                    The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
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                    Confectionery

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                    Prepared foods made with a lot of sugar or other cabohydrates "Sweetmeat" redirects here. For the racehorse, see Sweetmeat (horse). Not to be confused with Sweetbread. This Kransekake is a traditional Scandinavian baker's confection. Confectionery is the art of making confections , which are food items that are rich in sugar and carbohydrates. Exact definitions are difficult. [1] In general, though, confectionery is divided into two broad and somewhat overlapping categories, bakers' confections and sugar confections. [2] Bakers' confectionery, also called flour confections , includes principally sweet pastries, cakes, and similar baked goods. Sugar confectionery includes candies ( sweets in British English), candied nuts, chocolates, chewing gum, bubble gum, pastillage, and other confections that are made primarily of sugar. In some cases, chocolate confections (confections made of chocolate) are treated as a separate category, as are sugar-free versions of ...
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