volume of the solid generated by $y^{2}+z^{2}=1 ,y= frac{1}{2}$ and y greater than equal to $0$ about $z$...












0












$begingroup$


volume of the solid generated by $y^{2}+z^{2}=1$ ,$y= frac{1}{2}$ and $y$ greater than equal to 0 about $ z axis $



enter image description here



I have tried by disk method but able to get answer . I have got this figure enter image description here



i have tried breaking region from $z=frac{sqrt{3}}{2}$



$int_{0}^{frac{sqrt{3}}{2}}pi (frac{1}{2})^{2}dz$+$int_{frac{sqrt{3}}{2}}^{1}pi(sqrt{1-z^{2}})^{2}dz$



$=frac{sqrt{3}pi}{8}+frac{2}{3}+frac{3sqrt{3}}{8}$



$=frac{sqrt{3}pi}{4}+frac{2}{3}$



Please help .(this question was asked in IIT JAM2018)










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closed as off-topic by José Carlos Santos, Saad, Abcd, Leucippus, user91500 Jan 5 at 8:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, Abcd, Leucippus, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Please write explicitly your answer, including how you got there. There might be a typo in the question.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:05










  • $begingroup$
    no , the question is correct
    $endgroup$
    – sejy
    Dec 26 '18 at 16:40








  • 1




    $begingroup$
    Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:51










  • $begingroup$
    i have used disk method to solve this !! and answer key says c is the correct part
    $endgroup$
    – sejy
    Dec 26 '18 at 17:48


















0












$begingroup$


volume of the solid generated by $y^{2}+z^{2}=1$ ,$y= frac{1}{2}$ and $y$ greater than equal to 0 about $ z axis $



enter image description here



I have tried by disk method but able to get answer . I have got this figure enter image description here



i have tried breaking region from $z=frac{sqrt{3}}{2}$



$int_{0}^{frac{sqrt{3}}{2}}pi (frac{1}{2})^{2}dz$+$int_{frac{sqrt{3}}{2}}^{1}pi(sqrt{1-z^{2}})^{2}dz$



$=frac{sqrt{3}pi}{8}+frac{2}{3}+frac{3sqrt{3}}{8}$



$=frac{sqrt{3}pi}{4}+frac{2}{3}$



Please help .(this question was asked in IIT JAM2018)










share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, Saad, Abcd, Leucippus, user91500 Jan 5 at 8:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, Abcd, Leucippus, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Please write explicitly your answer, including how you got there. There might be a typo in the question.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:05










  • $begingroup$
    no , the question is correct
    $endgroup$
    – sejy
    Dec 26 '18 at 16:40








  • 1




    $begingroup$
    Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:51










  • $begingroup$
    i have used disk method to solve this !! and answer key says c is the correct part
    $endgroup$
    – sejy
    Dec 26 '18 at 17:48
















0












0








0


1



$begingroup$


volume of the solid generated by $y^{2}+z^{2}=1$ ,$y= frac{1}{2}$ and $y$ greater than equal to 0 about $ z axis $



enter image description here



I have tried by disk method but able to get answer . I have got this figure enter image description here



i have tried breaking region from $z=frac{sqrt{3}}{2}$



$int_{0}^{frac{sqrt{3}}{2}}pi (frac{1}{2})^{2}dz$+$int_{frac{sqrt{3}}{2}}^{1}pi(sqrt{1-z^{2}})^{2}dz$



$=frac{sqrt{3}pi}{8}+frac{2}{3}+frac{3sqrt{3}}{8}$



$=frac{sqrt{3}pi}{4}+frac{2}{3}$



Please help .(this question was asked in IIT JAM2018)










share|cite|improve this question











$endgroup$




volume of the solid generated by $y^{2}+z^{2}=1$ ,$y= frac{1}{2}$ and $y$ greater than equal to 0 about $ z axis $



enter image description here



I have tried by disk method but able to get answer . I have got this figure enter image description here



i have tried breaking region from $z=frac{sqrt{3}}{2}$



$int_{0}^{frac{sqrt{3}}{2}}pi (frac{1}{2})^{2}dz$+$int_{frac{sqrt{3}}{2}}^{1}pi(sqrt{1-z^{2}})^{2}dz$



$=frac{sqrt{3}pi}{8}+frac{2}{3}+frac{3sqrt{3}}{8}$



$=frac{sqrt{3}pi}{4}+frac{2}{3}$



Please help .(this question was asked in IIT JAM2018)







calculus multivariable-calculus multiple-integral






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edited Dec 26 '18 at 17:47







sejy

















asked Dec 26 '18 at 15:08









sejysejy

1539




1539




closed as off-topic by José Carlos Santos, Saad, Abcd, Leucippus, user91500 Jan 5 at 8:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, Abcd, Leucippus, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by José Carlos Santos, Saad, Abcd, Leucippus, user91500 Jan 5 at 8:05


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Saad, Abcd, Leucippus, user91500

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Please write explicitly your answer, including how you got there. There might be a typo in the question.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:05










  • $begingroup$
    no , the question is correct
    $endgroup$
    – sejy
    Dec 26 '18 at 16:40








  • 1




    $begingroup$
    Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:51










  • $begingroup$
    i have used disk method to solve this !! and answer key says c is the correct part
    $endgroup$
    – sejy
    Dec 26 '18 at 17:48
















  • 1




    $begingroup$
    Please write explicitly your answer, including how you got there. There might be a typo in the question.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:05










  • $begingroup$
    no , the question is correct
    $endgroup$
    – sejy
    Dec 26 '18 at 16:40








  • 1




    $begingroup$
    Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
    $endgroup$
    – Andrei
    Dec 26 '18 at 16:51










  • $begingroup$
    i have used disk method to solve this !! and answer key says c is the correct part
    $endgroup$
    – sejy
    Dec 26 '18 at 17:48










1




1




$begingroup$
Please write explicitly your answer, including how you got there. There might be a typo in the question.
$endgroup$
– Andrei
Dec 26 '18 at 16:05




$begingroup$
Please write explicitly your answer, including how you got there. There might be a typo in the question.
$endgroup$
– Andrei
Dec 26 '18 at 16:05












$begingroup$
no , the question is correct
$endgroup$
– sejy
Dec 26 '18 at 16:40






$begingroup$
no , the question is correct
$endgroup$
– sejy
Dec 26 '18 at 16:40






1




1




$begingroup$
Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
$endgroup$
– Andrei
Dec 26 '18 at 16:51




$begingroup$
Please show ALL the work you have done. Type in all the formulas you have used, and all intermediate answers. You might have done everything correctly, but the answer is not one of ABCD.
$endgroup$
– Andrei
Dec 26 '18 at 16:51












$begingroup$
i have used disk method to solve this !! and answer key says c is the correct part
$endgroup$
– sejy
Dec 26 '18 at 17:48






$begingroup$
i have used disk method to solve this !! and answer key says c is the correct part
$endgroup$
– sejy
Dec 26 '18 at 17:48












3 Answers
3






active

oldest

votes


















2












$begingroup$

enter image description here



Note that the strip $AB$ should be on the right of the vertical line $y = frac{1}{2}$



Consider the the elementary strip of width $dy$, whose end points are $A (y, sqrt{1-y^2})$ and $B(y, -sqrt{1-y^2})$



When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 pi y (2sqrt{1-y^2}) dy = (-2 pi) sqrt{1-y^2} (-2y) dy$



Hence the volume of the solid is $$int_{frac{1}{2}}^1(-2 pi) sqrt{1-y^2} (-2y) dy = left[frac{4 pi}{3}left(1-y^2right)^{frac{3}{2}}right]_1^{frac{1}{2}} = frac{pi sqrt{3}}{2}$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Your integral calculation is wrong. Let's start with that. First of all, you lost some $pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$frac V2=frac{2pi}3-frac{pisqrt 3}{4}$$ or $$V=frac{4pi}3-frac{pisqrt 3}{2}$$
    I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2sqrt{1-y^2}$. So the results are consistent.



    Now going back to the formula, the value from answer (C) is $frac{4pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $sqrt{1-z^2}$ $$V=int_{-sqrt 3/2}^{sqrt3/2}pi(1-z^2-frac1{2^2})dz$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      First, a few things that are wrong in your calculation.




      • In "$frac{sqrt{3}pi}{8}+color{red}{frac{2}{3}+frac{3sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$pi$" from the integral.


      • Then you somehow combined the two terms "$frac{sqrt{3}pi}{8}+cdots+frac{3sqrt{3}}{8}$" into "$color{red}{frac{sqrt{3}pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.





      Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=dfrac{1}{2}$, not to its left. See here:



      enter image description here



      Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-dfrac{sqrt{3}}{2}$ to $z=dfrac{sqrt{3}}{2}$, like this:
      $$int_{-sqrt{3}/2}^{sqrt{3}/2}pileft(sqrt{1-z^2}right)^2,dz-int_{-sqrt{3}/2}^{sqrt{3}/2}[text{the integral for the hole}].$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
        $endgroup$
        – Andrei
        Dec 26 '18 at 18:27


















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      enter image description here



      Note that the strip $AB$ should be on the right of the vertical line $y = frac{1}{2}$



      Consider the the elementary strip of width $dy$, whose end points are $A (y, sqrt{1-y^2})$ and $B(y, -sqrt{1-y^2})$



      When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 pi y (2sqrt{1-y^2}) dy = (-2 pi) sqrt{1-y^2} (-2y) dy$



      Hence the volume of the solid is $$int_{frac{1}{2}}^1(-2 pi) sqrt{1-y^2} (-2y) dy = left[frac{4 pi}{3}left(1-y^2right)^{frac{3}{2}}right]_1^{frac{1}{2}} = frac{pi sqrt{3}}{2}$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        enter image description here



        Note that the strip $AB$ should be on the right of the vertical line $y = frac{1}{2}$



        Consider the the elementary strip of width $dy$, whose end points are $A (y, sqrt{1-y^2})$ and $B(y, -sqrt{1-y^2})$



        When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 pi y (2sqrt{1-y^2}) dy = (-2 pi) sqrt{1-y^2} (-2y) dy$



        Hence the volume of the solid is $$int_{frac{1}{2}}^1(-2 pi) sqrt{1-y^2} (-2y) dy = left[frac{4 pi}{3}left(1-y^2right)^{frac{3}{2}}right]_1^{frac{1}{2}} = frac{pi sqrt{3}}{2}$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          enter image description here



          Note that the strip $AB$ should be on the right of the vertical line $y = frac{1}{2}$



          Consider the the elementary strip of width $dy$, whose end points are $A (y, sqrt{1-y^2})$ and $B(y, -sqrt{1-y^2})$



          When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 pi y (2sqrt{1-y^2}) dy = (-2 pi) sqrt{1-y^2} (-2y) dy$



          Hence the volume of the solid is $$int_{frac{1}{2}}^1(-2 pi) sqrt{1-y^2} (-2y) dy = left[frac{4 pi}{3}left(1-y^2right)^{frac{3}{2}}right]_1^{frac{1}{2}} = frac{pi sqrt{3}}{2}$$






          share|cite|improve this answer











          $endgroup$



          enter image description here



          Note that the strip $AB$ should be on the right of the vertical line $y = frac{1}{2}$



          Consider the the elementary strip of width $dy$, whose end points are $A (y, sqrt{1-y^2})$ and $B(y, -sqrt{1-y^2})$



          When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 pi y (2sqrt{1-y^2}) dy = (-2 pi) sqrt{1-y^2} (-2y) dy$



          Hence the volume of the solid is $$int_{frac{1}{2}}^1(-2 pi) sqrt{1-y^2} (-2y) dy = left[frac{4 pi}{3}left(1-y^2right)^{frac{3}{2}}right]_1^{frac{1}{2}} = frac{pi sqrt{3}}{2}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 5:14

























          answered Dec 27 '18 at 2:46









          PTDSPTDS

          56923




          56923























              2












              $begingroup$

              Your integral calculation is wrong. Let's start with that. First of all, you lost some $pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$frac V2=frac{2pi}3-frac{pisqrt 3}{4}$$ or $$V=frac{4pi}3-frac{pisqrt 3}{2}$$
              I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2sqrt{1-y^2}$. So the results are consistent.



              Now going back to the formula, the value from answer (C) is $frac{4pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $sqrt{1-z^2}$ $$V=int_{-sqrt 3/2}^{sqrt3/2}pi(1-z^2-frac1{2^2})dz$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Your integral calculation is wrong. Let's start with that. First of all, you lost some $pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$frac V2=frac{2pi}3-frac{pisqrt 3}{4}$$ or $$V=frac{4pi}3-frac{pisqrt 3}{2}$$
                I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2sqrt{1-y^2}$. So the results are consistent.



                Now going back to the formula, the value from answer (C) is $frac{4pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $sqrt{1-z^2}$ $$V=int_{-sqrt 3/2}^{sqrt3/2}pi(1-z^2-frac1{2^2})dz$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Your integral calculation is wrong. Let's start with that. First of all, you lost some $pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$frac V2=frac{2pi}3-frac{pisqrt 3}{4}$$ or $$V=frac{4pi}3-frac{pisqrt 3}{2}$$
                  I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2sqrt{1-y^2}$. So the results are consistent.



                  Now going back to the formula, the value from answer (C) is $frac{4pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $sqrt{1-z^2}$ $$V=int_{-sqrt 3/2}^{sqrt3/2}pi(1-z^2-frac1{2^2})dz$$






                  share|cite|improve this answer









                  $endgroup$



                  Your integral calculation is wrong. Let's start with that. First of all, you lost some $pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$frac V2=frac{2pi}3-frac{pisqrt 3}{4}$$ or $$V=frac{4pi}3-frac{pisqrt 3}{2}$$
                  I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2sqrt{1-y^2}$. So the results are consistent.



                  Now going back to the formula, the value from answer (C) is $frac{4pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $sqrt{1-z^2}$ $$V=int_{-sqrt 3/2}^{sqrt3/2}pi(1-z^2-frac1{2^2})dz$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 26 '18 at 18:20









                  AndreiAndrei

                  12.4k21128




                  12.4k21128























                      2












                      $begingroup$

                      First, a few things that are wrong in your calculation.




                      • In "$frac{sqrt{3}pi}{8}+color{red}{frac{2}{3}+frac{3sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$pi$" from the integral.


                      • Then you somehow combined the two terms "$frac{sqrt{3}pi}{8}+cdots+frac{3sqrt{3}}{8}$" into "$color{red}{frac{sqrt{3}pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.





                      Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=dfrac{1}{2}$, not to its left. See here:



                      enter image description here



                      Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-dfrac{sqrt{3}}{2}$ to $z=dfrac{sqrt{3}}{2}$, like this:
                      $$int_{-sqrt{3}/2}^{sqrt{3}/2}pileft(sqrt{1-z^2}right)^2,dz-int_{-sqrt{3}/2}^{sqrt{3}/2}[text{the integral for the hole}].$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                        $endgroup$
                        – Andrei
                        Dec 26 '18 at 18:27
















                      2












                      $begingroup$

                      First, a few things that are wrong in your calculation.




                      • In "$frac{sqrt{3}pi}{8}+color{red}{frac{2}{3}+frac{3sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$pi$" from the integral.


                      • Then you somehow combined the two terms "$frac{sqrt{3}pi}{8}+cdots+frac{3sqrt{3}}{8}$" into "$color{red}{frac{sqrt{3}pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.





                      Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=dfrac{1}{2}$, not to its left. See here:



                      enter image description here



                      Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-dfrac{sqrt{3}}{2}$ to $z=dfrac{sqrt{3}}{2}$, like this:
                      $$int_{-sqrt{3}/2}^{sqrt{3}/2}pileft(sqrt{1-z^2}right)^2,dz-int_{-sqrt{3}/2}^{sqrt{3}/2}[text{the integral for the hole}].$$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                        $endgroup$
                        – Andrei
                        Dec 26 '18 at 18:27














                      2












                      2








                      2





                      $begingroup$

                      First, a few things that are wrong in your calculation.




                      • In "$frac{sqrt{3}pi}{8}+color{red}{frac{2}{3}+frac{3sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$pi$" from the integral.


                      • Then you somehow combined the two terms "$frac{sqrt{3}pi}{8}+cdots+frac{3sqrt{3}}{8}$" into "$color{red}{frac{sqrt{3}pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.





                      Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=dfrac{1}{2}$, not to its left. See here:



                      enter image description here



                      Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-dfrac{sqrt{3}}{2}$ to $z=dfrac{sqrt{3}}{2}$, like this:
                      $$int_{-sqrt{3}/2}^{sqrt{3}/2}pileft(sqrt{1-z^2}right)^2,dz-int_{-sqrt{3}/2}^{sqrt{3}/2}[text{the integral for the hole}].$$






                      share|cite|improve this answer









                      $endgroup$



                      First, a few things that are wrong in your calculation.




                      • In "$frac{sqrt{3}pi}{8}+color{red}{frac{2}{3}+frac{3sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$pi$" from the integral.


                      • Then you somehow combined the two terms "$frac{sqrt{3}pi}{8}+cdots+frac{3sqrt{3}}{8}$" into "$color{red}{frac{sqrt{3}pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.





                      Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=dfrac{1}{2}$, not to its left. See here:



                      enter image description here



                      Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-dfrac{sqrt{3}}{2}$ to $z=dfrac{sqrt{3}}{2}$, like this:
                      $$int_{-sqrt{3}/2}^{sqrt{3}/2}pileft(sqrt{1-z^2}right)^2,dz-int_{-sqrt{3}/2}^{sqrt{3}/2}[text{the integral for the hole}].$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 26 '18 at 18:22









                      zipirovichzipirovich

                      11.3k11731




                      11.3k11731












                      • $begingroup$
                        The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                        $endgroup$
                        – Andrei
                        Dec 26 '18 at 18:27


















                      • $begingroup$
                        The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                        $endgroup$
                        – Andrei
                        Dec 26 '18 at 18:27
















                      $begingroup$
                      The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                      $endgroup$
                      – Andrei
                      Dec 26 '18 at 18:27




                      $begingroup$
                      The $yge 0$ was confusing for me as well. It took me a while to guess what they wanted, and I am still be in doubt if their sentence makes sense
                      $endgroup$
                      – Andrei
                      Dec 26 '18 at 18:27



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