Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.

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Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.



I know how to graph it out but I have no idea what to do after that.










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    Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.



    I know how to graph it out but I have no idea what to do after that.










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      Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.



      I know how to graph it out but I have no idea what to do after that.










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      Find the number b such that the line y=b divides the region bounded by y=0 and y=4-x^2 into two regions with equal area.



      I know how to graph it out but I have no idea what to do after that.







      calculus






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      asked Apr 26 '17 at 3:40









      tortugaaaaaaaa

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          The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
          We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation



          $$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
          and ultimately to $b=colorred2(2-sqrt[3]2)$.






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            You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
            $$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
            and you can proceeded here.






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              The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
              We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation



              $$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
              and ultimately to $b=colorred2(2-sqrt[3]2)$.






              share|cite|improve this answer
























                up vote
                0
                down vote













                The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
                We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation



                $$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
                and ultimately to $b=colorred2(2-sqrt[3]2)$.






                share|cite|improve this answer






















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                  up vote
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                  The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
                  We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation



                  $$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
                  and ultimately to $b=colorred2(2-sqrt[3]2)$.






                  share|cite|improve this answer












                  The area between $y=4-x^2$ and $y=0$ is $int_-2^2(4-x^2),dx = frac323$.
                  We have to find some constant $bin(0,4)$ such that the area between $y=4-x^2$ and $y=b$ equals $frac163$. That leads to the equation



                  $$ int_-sqrt4-b^sqrt4-b(4-b-x^2),dx = frac163 $$
                  and ultimately to $b=colorred2(2-sqrt[3]2)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 26 '17 at 3:47









                  Jack D'Aurizio♦

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                      0
                      down vote













                      You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
                      $$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
                      and you can proceeded here.






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
                        $$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
                        and you can proceeded here.






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
                          $$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
                          and you can proceeded here.






                          share|cite|improve this answer












                          You know how to graph it out and you find it symmetric about $y$-axis. So we can let the problem to right side of $y$-axis. The integrals in intervals $[0,b]$ and $[b,4]$ respect to $y$-axis are
                          $$int_0^bsqrt4-y,dy=int_b^4sqrt4-y,dy$$
                          and you can proceeded here.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 26 '17 at 3:51









                          Nosrati

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