Area enclosed by polar curve












4














I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer. Can you assist in clarification?



Find the area of the region enclosed by $r=4cos(3 theta)$.



I use $ frac 12 int_0^{2pi} (16cos^2(3theta) dtheta$. For $cos^2(3theta)$ I use the identity $frac12[1+cos(6theta)]$



This gives me $frac{16}{4} int_0^{2pi} 1+cos(6theta) dtheta$.



This gives me $4[int_0^{2pi}1 dtheta +frac16int_0^{12pi} cos (u) du]$.



The integral of the cosine term is $0$, so I get $theta $ evaluated from $0$ to $2pi$. This gives me $4(2pi)=8pi$.



When I use a symmetrical method A=$6int_0^{pi/6}frac12(16 cos^2(3theta)dtheta$ I get $4pi$. This is the text answer.



Don't understand why my 2 answers don't match.










share|cite|improve this question



























    4














    I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer. Can you assist in clarification?



    Find the area of the region enclosed by $r=4cos(3 theta)$.



    I use $ frac 12 int_0^{2pi} (16cos^2(3theta) dtheta$. For $cos^2(3theta)$ I use the identity $frac12[1+cos(6theta)]$



    This gives me $frac{16}{4} int_0^{2pi} 1+cos(6theta) dtheta$.



    This gives me $4[int_0^{2pi}1 dtheta +frac16int_0^{12pi} cos (u) du]$.



    The integral of the cosine term is $0$, so I get $theta $ evaluated from $0$ to $2pi$. This gives me $4(2pi)=8pi$.



    When I use a symmetrical method A=$6int_0^{pi/6}frac12(16 cos^2(3theta)dtheta$ I get $4pi$. This is the text answer.



    Don't understand why my 2 answers don't match.










    share|cite|improve this question

























      4












      4








      4







      I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer. Can you assist in clarification?



      Find the area of the region enclosed by $r=4cos(3 theta)$.



      I use $ frac 12 int_0^{2pi} (16cos^2(3theta) dtheta$. For $cos^2(3theta)$ I use the identity $frac12[1+cos(6theta)]$



      This gives me $frac{16}{4} int_0^{2pi} 1+cos(6theta) dtheta$.



      This gives me $4[int_0^{2pi}1 dtheta +frac16int_0^{12pi} cos (u) du]$.



      The integral of the cosine term is $0$, so I get $theta $ evaluated from $0$ to $2pi$. This gives me $4(2pi)=8pi$.



      When I use a symmetrical method A=$6int_0^{pi/6}frac12(16 cos^2(3theta)dtheta$ I get $4pi$. This is the text answer.



      Don't understand why my 2 answers don't match.










      share|cite|improve this question













      I can't get the text answer using standard method of integration of a polar equation. Yet when I use a symmetry method I do get their answer. Can you assist in clarification?



      Find the area of the region enclosed by $r=4cos(3 theta)$.



      I use $ frac 12 int_0^{2pi} (16cos^2(3theta) dtheta$. For $cos^2(3theta)$ I use the identity $frac12[1+cos(6theta)]$



      This gives me $frac{16}{4} int_0^{2pi} 1+cos(6theta) dtheta$.



      This gives me $4[int_0^{2pi}1 dtheta +frac16int_0^{12pi} cos (u) du]$.



      The integral of the cosine term is $0$, so I get $theta $ evaluated from $0$ to $2pi$. This gives me $4(2pi)=8pi$.



      When I use a symmetrical method A=$6int_0^{pi/6}frac12(16 cos^2(3theta)dtheta$ I get $4pi$. This is the text answer.



      Don't understand why my 2 answers don't match.







      calculus






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      asked Dec 26 '18 at 18:30









      user163862

      84821016




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          3 Answers
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          4














          Your first answer is twice the correct answer for the following reason: if you let $theta$ range from $theta=0$ to $theta=2pi$, the curve $r=4cos(3theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $theta=pi$; and then, from $theta=pi$ to $theta=2pi$ you retrace it once more.



          In the second method, you find the area of a half of one petal, which you correctly determined to range from $theta=0$ to $theta=dfrac{pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $theta=0$ to $theta=dfrac{pi}{6}$ will produce the angle six times as wide, i.e. from $theta=0$ to $theta=pi$, consistent with my explanation above.






          share|cite|improve this answer























          • Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
            – user163862
            Dec 26 '18 at 20:05



















          -1














          $$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$






          share|cite|improve this answer





























            -1














            To provide yet another approach, you can use Green's theorem on the differential form $x ,dy$ to transform the area integral over one petal into a line integral along the petal:



            $$frac13 A = text{area of petal} = intlimits_{text{petal}},1,dA = intlimits_{text{boundary of petal}} x,dy $$



            Since $r = 4cos(3theta)$, the petal can be parameterized by
            $$(x(theta), y(theta)) = (r(theta)costheta, r(theta)sin(theta)) = (4costhetacos(3theta), 4sinthetacos(3theta))$$



            for $theta in left[-frac{pi}6, frac{pi}6right]$.



            Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so



            $$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$



            so $A = 4pi$.



            The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.






            share|cite|improve this answer























            • ... and which is the wrong answer of the two.
              – zipirovich
              Dec 26 '18 at 18:51










            • @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
              – mechanodroid
              Dec 26 '18 at 19:36











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            Your first answer is twice the correct answer for the following reason: if you let $theta$ range from $theta=0$ to $theta=2pi$, the curve $r=4cos(3theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $theta=pi$; and then, from $theta=pi$ to $theta=2pi$ you retrace it once more.



            In the second method, you find the area of a half of one petal, which you correctly determined to range from $theta=0$ to $theta=dfrac{pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $theta=0$ to $theta=dfrac{pi}{6}$ will produce the angle six times as wide, i.e. from $theta=0$ to $theta=pi$, consistent with my explanation above.






            share|cite|improve this answer























            • Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
              – user163862
              Dec 26 '18 at 20:05
















            4














            Your first answer is twice the correct answer for the following reason: if you let $theta$ range from $theta=0$ to $theta=2pi$, the curve $r=4cos(3theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $theta=pi$; and then, from $theta=pi$ to $theta=2pi$ you retrace it once more.



            In the second method, you find the area of a half of one petal, which you correctly determined to range from $theta=0$ to $theta=dfrac{pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $theta=0$ to $theta=dfrac{pi}{6}$ will produce the angle six times as wide, i.e. from $theta=0$ to $theta=pi$, consistent with my explanation above.






            share|cite|improve this answer























            • Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
              – user163862
              Dec 26 '18 at 20:05














            4












            4








            4






            Your first answer is twice the correct answer for the following reason: if you let $theta$ range from $theta=0$ to $theta=2pi$, the curve $r=4cos(3theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $theta=pi$; and then, from $theta=pi$ to $theta=2pi$ you retrace it once more.



            In the second method, you find the area of a half of one petal, which you correctly determined to range from $theta=0$ to $theta=dfrac{pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $theta=0$ to $theta=dfrac{pi}{6}$ will produce the angle six times as wide, i.e. from $theta=0$ to $theta=pi$, consistent with my explanation above.






            share|cite|improve this answer














            Your first answer is twice the correct answer for the following reason: if you let $theta$ range from $theta=0$ to $theta=2pi$, the curve $r=4cos(3theta)$ — which is a flower with three petals — is traced twice, and therefore you find twice the area. If you trace it carefully starting from $theta=0$, which is $(4,0)$ in cartesian coordinates, you will see that the curve is completed and comes back to the initial point at $theta=pi$; and then, from $theta=pi$ to $theta=2pi$ you retrace it once more.



            In the second method, you find the area of a half of one petal, which you correctly determined to range from $theta=0$ to $theta=dfrac{pi}{6}$. Since there are six such half-petals, multiplying by $6$ clearly yields the correct answer. Note, however, that taking six half-petals of the same "angular width" (so to speak) as the one going from $theta=0$ to $theta=dfrac{pi}{6}$ will produce the angle six times as wide, i.e. from $theta=0$ to $theta=pi$, consistent with my explanation above.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 26 '18 at 19:46

























            answered Dec 26 '18 at 18:49









            zipirovich

            10.9k11631




            10.9k11631












            • Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
              – user163862
              Dec 26 '18 at 20:05


















            • Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
              – user163862
              Dec 26 '18 at 20:05
















            Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
            – user163862
            Dec 26 '18 at 20:05




            Ah, I immediately see this now. How can you tell when you have a polar equation that traces out completely in $0$ to $pi$ instead of $0$ to $2pi$?
            – user163862
            Dec 26 '18 at 20:05











            -1














            $$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$






            share|cite|improve this answer


























              -1














              $$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$






              share|cite|improve this answer
























                -1












                -1








                -1






                $$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$






                share|cite|improve this answer












                $$intlimits_{theta = 0}^{2 pi} 4 |cos (3 theta)| dtheta = 16$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 26 '18 at 18:58









                David G. Stork

                9,82021232




                9,82021232























                    -1














                    To provide yet another approach, you can use Green's theorem on the differential form $x ,dy$ to transform the area integral over one petal into a line integral along the petal:



                    $$frac13 A = text{area of petal} = intlimits_{text{petal}},1,dA = intlimits_{text{boundary of petal}} x,dy $$



                    Since $r = 4cos(3theta)$, the petal can be parameterized by
                    $$(x(theta), y(theta)) = (r(theta)costheta, r(theta)sin(theta)) = (4costhetacos(3theta), 4sinthetacos(3theta))$$



                    for $theta in left[-frac{pi}6, frac{pi}6right]$.



                    Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so



                    $$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$



                    so $A = 4pi$.



                    The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.






                    share|cite|improve this answer























                    • ... and which is the wrong answer of the two.
                      – zipirovich
                      Dec 26 '18 at 18:51










                    • @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                      – mechanodroid
                      Dec 26 '18 at 19:36
















                    -1














                    To provide yet another approach, you can use Green's theorem on the differential form $x ,dy$ to transform the area integral over one petal into a line integral along the petal:



                    $$frac13 A = text{area of petal} = intlimits_{text{petal}},1,dA = intlimits_{text{boundary of petal}} x,dy $$



                    Since $r = 4cos(3theta)$, the petal can be parameterized by
                    $$(x(theta), y(theta)) = (r(theta)costheta, r(theta)sin(theta)) = (4costhetacos(3theta), 4sinthetacos(3theta))$$



                    for $theta in left[-frac{pi}6, frac{pi}6right]$.



                    Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so



                    $$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$



                    so $A = 4pi$.



                    The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.






                    share|cite|improve this answer























                    • ... and which is the wrong answer of the two.
                      – zipirovich
                      Dec 26 '18 at 18:51










                    • @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                      – mechanodroid
                      Dec 26 '18 at 19:36














                    -1












                    -1








                    -1






                    To provide yet another approach, you can use Green's theorem on the differential form $x ,dy$ to transform the area integral over one petal into a line integral along the petal:



                    $$frac13 A = text{area of petal} = intlimits_{text{petal}},1,dA = intlimits_{text{boundary of petal}} x,dy $$



                    Since $r = 4cos(3theta)$, the petal can be parameterized by
                    $$(x(theta), y(theta)) = (r(theta)costheta, r(theta)sin(theta)) = (4costhetacos(3theta), 4sinthetacos(3theta))$$



                    for $theta in left[-frac{pi}6, frac{pi}6right]$.



                    Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so



                    $$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$



                    so $A = 4pi$.



                    The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.






                    share|cite|improve this answer














                    To provide yet another approach, you can use Green's theorem on the differential form $x ,dy$ to transform the area integral over one petal into a line integral along the petal:



                    $$frac13 A = text{area of petal} = intlimits_{text{petal}},1,dA = intlimits_{text{boundary of petal}} x,dy $$



                    Since $r = 4cos(3theta)$, the petal can be parameterized by
                    $$(x(theta), y(theta)) = (r(theta)costheta, r(theta)sin(theta)) = (4costhetacos(3theta), 4sinthetacos(3theta))$$



                    for $theta in left[-frac{pi}6, frac{pi}6right]$.



                    Differentiating gives $dy = (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta$ so



                    $$frac13 A = int_{-frac{pi}6}^{frac{pi}6}4costhetacos(3theta) (4costhetacos(3theta) - 12sinthetasin(3theta)),dtheta = frac{4pi}3$$



                    so $A = 4pi$.



                    The integral is a bit cumbersome but it can be solved using the product-to-sum formulas for sine and cosine.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 26 '18 at 19:35

























                    answered Dec 26 '18 at 18:40









                    mechanodroid

                    26.1k62245




                    26.1k62245












                    • ... and which is the wrong answer of the two.
                      – zipirovich
                      Dec 26 '18 at 18:51










                    • @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                      – mechanodroid
                      Dec 26 '18 at 19:36


















                    • ... and which is the wrong answer of the two.
                      – zipirovich
                      Dec 26 '18 at 18:51










                    • @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                      – mechanodroid
                      Dec 26 '18 at 19:36
















                    ... and which is the wrong answer of the two.
                    – zipirovich
                    Dec 26 '18 at 18:51




                    ... and which is the wrong answer of the two.
                    – zipirovich
                    Dec 26 '18 at 18:51












                    @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                    – mechanodroid
                    Dec 26 '18 at 19:36




                    @zipirovich You are right of course, thanks. I have added a completely different approach to obtain the right result.
                    – mechanodroid
                    Dec 26 '18 at 19:36


















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