Setting up integrals for the moments and the center of mass of a planar region












3












$begingroup$


Problem



Consider the following region: a semi-circle with radius = 3 ft on top of a rectangle with height = 11. (with constant density)



a.) Set up integrals for the moments, Mx, My, and the center of mass of the region. DO NOT evaluate the integrals.



b.) Use additivity of moments to find the center of mass of the region.



Progress



For part a.) I'm assuming I have to find separate Mx and My for both the semi-circle and the rectangle and add the Mx of semi-circle and Mx of rectangle together. Is that the correct way to go at this problem?



For the semi-circle $$M_x = int _{-3}^3:frac{1}{2}left(sqrt{3^2-x^2}right)^2 dx$$ $M_y= 0$ because of symmetry and constant density.



$M_i= 3pi $ for the rectangle I think I have to find out what $fleft(xright)$ is. How would I do this?






calculus integration







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    3












    $begingroup$


    Problem



    Consider the following region: a semi-circle with radius = 3 ft on top of a rectangle with height = 11. (with constant density)



    a.) Set up integrals for the moments, Mx, My, and the center of mass of the region. DO NOT evaluate the integrals.



    b.) Use additivity of moments to find the center of mass of the region.



    Progress



    For part a.) I'm assuming I have to find separate Mx and My for both the semi-circle and the rectangle and add the Mx of semi-circle and Mx of rectangle together. Is that the correct way to go at this problem?



    For the semi-circle $$M_x = int _{-3}^3:frac{1}{2}left(sqrt{3^2-x^2}right)^2 dx$$ $M_y= 0$ because of symmetry and constant density.



    $M_i= 3pi $ for the rectangle I think I have to find out what $fleft(xright)$ is. How would I do this?






    calculus integration







    share|cite|improve this question











    $endgroup$




    bumped to the homepage by Community♦ yesterday


    This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.


















      3












      3








      3


      1



      $begingroup$


      Problem



      Consider the following region: a semi-circle with radius = 3 ft on top of a rectangle with height = 11. (with constant density)



      a.) Set up integrals for the moments, Mx, My, and the center of mass of the region. DO NOT evaluate the integrals.



      b.) Use additivity of moments to find the center of mass of the region.



      Progress



      For part a.) I'm assuming I have to find separate Mx and My for both the semi-circle and the rectangle and add the Mx of semi-circle and Mx of rectangle together. Is that the correct way to go at this problem?



      For the semi-circle $$M_x = int _{-3}^3:frac{1}{2}left(sqrt{3^2-x^2}right)^2 dx$$ $M_y= 0$ because of symmetry and constant density.



      $M_i= 3pi $ for the rectangle I think I have to find out what $fleft(xright)$ is. How would I do this?






      calculus integration







      share|cite|improve this question











      $endgroup$




      Problem



      Consider the following region: a semi-circle with radius = 3 ft on top of a rectangle with height = 11. (with constant density)



      a.) Set up integrals for the moments, Mx, My, and the center of mass of the region. DO NOT evaluate the integrals.



      b.) Use additivity of moments to find the center of mass of the region.



      Progress



      For part a.) I'm assuming I have to find separate Mx and My for both the semi-circle and the rectangle and add the Mx of semi-circle and Mx of rectangle together. Is that the correct way to go at this problem?



      For the semi-circle $$M_x = int _{-3}^3:frac{1}{2}left(sqrt{3^2-x^2}right)^2 dx$$ $M_y= 0$ because of symmetry and constant density.



      $M_i= 3pi $ for the rectangle I think I have to find out what $fleft(xright)$ is. How would I do this?






      calculus integration




      calculus integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Oct 20 '14 at 3:42







      user147263

















      asked Oct 20 '14 at 0:54









      CharleneCharlene

      402213




      402213





      bumped to the homepage by Community♦ yesterday


      This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







      bumped to the homepage by Community♦ yesterday


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          $begingroup$

          Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.



          If we take the density $delta=1$, we have that



          $displaystyle M_x=int_{-3}^{3}frac{1}{2}left((sqrt{9-x^2})^2-(-11)^2right)dx=-frac{1}{2}int_{-3}^{3}left(112+x^2right)dx$,



          $displaystyle M_y=int_{-3}^{3}xleft(sqrt{9-x^2}-(-11)right)dx=0$,



          so $displaystyleoverline{x}=frac{M_y}{m}=0$ $;;;$ and $;;;$$displaystyleoverline{y}=frac{M_x}{m}=frac{-frac{1}{2}int_{-3}^{3}(112+x^2)dx}{int_{-3}^{3}left(sqrt{9-x^2}-(-11)right)dx}$.




          For part b), we can use



          $displaystyle M_x=frac{1}{2}int_{-3}^{3}(9-x^2)dx=int_0^3(9-x^2)dx=left[9x-frac{x^3}{3}right]_0^3=18$ for the semicircle



          and $M_x=dm=(-frac{11}{2})(11)(6)=-363$ for the rectangle; so



          $;;;displaystyleoverline{y}=frac{M_x}{m}=frac{-363+18}{frac{1}{2}pi(3)^2+6(11)}=-frac{230}{3pi+43}$






          share|cite|improve this answer









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            $begingroup$

            Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.



            If we take the density $delta=1$, we have that



            $displaystyle M_x=int_{-3}^{3}frac{1}{2}left((sqrt{9-x^2})^2-(-11)^2right)dx=-frac{1}{2}int_{-3}^{3}left(112+x^2right)dx$,



            $displaystyle M_y=int_{-3}^{3}xleft(sqrt{9-x^2}-(-11)right)dx=0$,



            so $displaystyleoverline{x}=frac{M_y}{m}=0$ $;;;$ and $;;;$$displaystyleoverline{y}=frac{M_x}{m}=frac{-frac{1}{2}int_{-3}^{3}(112+x^2)dx}{int_{-3}^{3}left(sqrt{9-x^2}-(-11)right)dx}$.




            For part b), we can use



            $displaystyle M_x=frac{1}{2}int_{-3}^{3}(9-x^2)dx=int_0^3(9-x^2)dx=left[9x-frac{x^3}{3}right]_0^3=18$ for the semicircle



            and $M_x=dm=(-frac{11}{2})(11)(6)=-363$ for the rectangle; so



            $;;;displaystyleoverline{y}=frac{M_x}{m}=frac{-363+18}{frac{1}{2}pi(3)^2+6(11)}=-frac{230}{3pi+43}$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.



              If we take the density $delta=1$, we have that



              $displaystyle M_x=int_{-3}^{3}frac{1}{2}left((sqrt{9-x^2})^2-(-11)^2right)dx=-frac{1}{2}int_{-3}^{3}left(112+x^2right)dx$,



              $displaystyle M_y=int_{-3}^{3}xleft(sqrt{9-x^2}-(-11)right)dx=0$,



              so $displaystyleoverline{x}=frac{M_y}{m}=0$ $;;;$ and $;;;$$displaystyleoverline{y}=frac{M_x}{m}=frac{-frac{1}{2}int_{-3}^{3}(112+x^2)dx}{int_{-3}^{3}left(sqrt{9-x^2}-(-11)right)dx}$.




              For part b), we can use



              $displaystyle M_x=frac{1}{2}int_{-3}^{3}(9-x^2)dx=int_0^3(9-x^2)dx=left[9x-frac{x^3}{3}right]_0^3=18$ for the semicircle



              and $M_x=dm=(-frac{11}{2})(11)(6)=-363$ for the rectangle; so



              $;;;displaystyleoverline{y}=frac{M_x}{m}=frac{-363+18}{frac{1}{2}pi(3)^2+6(11)}=-frac{230}{3pi+43}$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.



                If we take the density $delta=1$, we have that



                $displaystyle M_x=int_{-3}^{3}frac{1}{2}left((sqrt{9-x^2})^2-(-11)^2right)dx=-frac{1}{2}int_{-3}^{3}left(112+x^2right)dx$,



                $displaystyle M_y=int_{-3}^{3}xleft(sqrt{9-x^2}-(-11)right)dx=0$,



                so $displaystyleoverline{x}=frac{M_y}{m}=0$ $;;;$ and $;;;$$displaystyleoverline{y}=frac{M_x}{m}=frac{-frac{1}{2}int_{-3}^{3}(112+x^2)dx}{int_{-3}^{3}left(sqrt{9-x^2}-(-11)right)dx}$.




                For part b), we can use



                $displaystyle M_x=frac{1}{2}int_{-3}^{3}(9-x^2)dx=int_0^3(9-x^2)dx=left[9x-frac{x^3}{3}right]_0^3=18$ for the semicircle



                and $M_x=dm=(-frac{11}{2})(11)(6)=-363$ for the rectangle; so



                $;;;displaystyleoverline{y}=frac{M_x}{m}=frac{-363+18}{frac{1}{2}pi(3)^2+6(11)}=-frac{230}{3pi+43}$






                share|cite|improve this answer









                $endgroup$



                Let's set up the coordinate axes so that the origin is at the center of the diameter of the semicircle, so the semicircle is given by $y=sqrt{9-x^2}$ as you have, and the bottom of the rectangle is given by $y=-11$.



                If we take the density $delta=1$, we have that



                $displaystyle M_x=int_{-3}^{3}frac{1}{2}left((sqrt{9-x^2})^2-(-11)^2right)dx=-frac{1}{2}int_{-3}^{3}left(112+x^2right)dx$,



                $displaystyle M_y=int_{-3}^{3}xleft(sqrt{9-x^2}-(-11)right)dx=0$,



                so $displaystyleoverline{x}=frac{M_y}{m}=0$ $;;;$ and $;;;$$displaystyleoverline{y}=frac{M_x}{m}=frac{-frac{1}{2}int_{-3}^{3}(112+x^2)dx}{int_{-3}^{3}left(sqrt{9-x^2}-(-11)right)dx}$.




                For part b), we can use



                $displaystyle M_x=frac{1}{2}int_{-3}^{3}(9-x^2)dx=int_0^3(9-x^2)dx=left[9x-frac{x^3}{3}right]_0^3=18$ for the semicircle



                and $M_x=dm=(-frac{11}{2})(11)(6)=-363$ for the rectangle; so



                $;;;displaystyleoverline{y}=frac{M_x}{m}=frac{-363+18}{frac{1}{2}pi(3)^2+6(11)}=-frac{230}{3pi+43}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 20 '14 at 22:49









                user84413user84413

                23k11848




                23k11848






























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