Finding volume of a region in Multi-Variable Calculus Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is the volume above the cone $z= sqrtx^2+y^2$ and bounded by the spheres $𝑥^2+y^2+𝑧^2=1$ and $𝑥^2+y^2+𝑧^2=4$?Single variable integral to polar coordinates?Finding a volumeVolume under hyperbolic paraboloid, above unit diskbasic question - Volume of revolution of weird shapeFinding the volume of the following solid using triple integralsVolume Integration of Bounded RegionFind the volume of the solid region given some constraints.Non translation polar Limits of integration for volume of cylinder not centered at originfinding volume bound by cylinder, cone, and xy planefind volume using double integral

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Finding volume of a region in Multi-Variable Calculus



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)What is the volume above the cone $z= sqrtx^2+y^2$ and bounded by the spheres $𝑥^2+y^2+𝑧^2=1$ and $𝑥^2+y^2+𝑧^2=4$?Single variable integral to polar coordinates?Finding a volumeVolume under hyperbolic paraboloid, above unit diskbasic question - Volume of revolution of weird shapeFinding the volume of the following solid using triple integralsVolume Integration of Bounded RegionFind the volume of the solid region given some constraints.Non translation polar Limits of integration for volume of cylinder not centered at originfinding volume bound by cylinder, cone, and xy planefind volume using double integral










0












$begingroup$


enter image description here



I converted both functions into polar form and got $z = r$ and $z = r^2 - rcostheta$. So I get the integrands as $[r^2]drdtheta$ and $[r^3 - r^2 costheta] drdtheta$ but now I'm having difficulty visualizing the shapes and setting the bounds on $r$ and $theta$. I assume $theta$ is from $0$ to $2π$ (correct me if I'm wrong) - but I still don't completely understand why this is the case. Also, I'm having trouble setting up $r$ bounds. When I choose $0<r<2$ I get a different volume for the cylinder than $2<r<4$ and none of them equal $$V= pi r^2h= 4pi(4) = 16pi.$$ Any help is appreciated. Thanks.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    enter image description here



    I converted both functions into polar form and got $z = r$ and $z = r^2 - rcostheta$. So I get the integrands as $[r^2]drdtheta$ and $[r^3 - r^2 costheta] drdtheta$ but now I'm having difficulty visualizing the shapes and setting the bounds on $r$ and $theta$. I assume $theta$ is from $0$ to $2π$ (correct me if I'm wrong) - but I still don't completely understand why this is the case. Also, I'm having trouble setting up $r$ bounds. When I choose $0<r<2$ I get a different volume for the cylinder than $2<r<4$ and none of them equal $$V= pi r^2h= 4pi(4) = 16pi.$$ Any help is appreciated. Thanks.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      enter image description here



      I converted both functions into polar form and got $z = r$ and $z = r^2 - rcostheta$. So I get the integrands as $[r^2]drdtheta$ and $[r^3 - r^2 costheta] drdtheta$ but now I'm having difficulty visualizing the shapes and setting the bounds on $r$ and $theta$. I assume $theta$ is from $0$ to $2π$ (correct me if I'm wrong) - but I still don't completely understand why this is the case. Also, I'm having trouble setting up $r$ bounds. When I choose $0<r<2$ I get a different volume for the cylinder than $2<r<4$ and none of them equal $$V= pi r^2h= 4pi(4) = 16pi.$$ Any help is appreciated. Thanks.










      share|cite|improve this question











      $endgroup$




      enter image description here



      I converted both functions into polar form and got $z = r$ and $z = r^2 - rcostheta$. So I get the integrands as $[r^2]drdtheta$ and $[r^3 - r^2 costheta] drdtheta$ but now I'm having difficulty visualizing the shapes and setting the bounds on $r$ and $theta$. I assume $theta$ is from $0$ to $2π$ (correct me if I'm wrong) - but I still don't completely understand why this is the case. Also, I'm having trouble setting up $r$ bounds. When I choose $0<r<2$ I get a different volume for the cylinder than $2<r<4$ and none of them equal $$V= pi r^2h= 4pi(4) = 16pi.$$ Any help is appreciated. Thanks.







      integration multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 27 at 20:09









      Gerschgorin

      8210




      8210










      asked Mar 27 at 19:37









      krauser126krauser126

      636




      636




















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          First - no, that's not the correct equation for the cylinder. There's no $z$ in the cylinder's equation; it should be $(rcostheta-2)^2+(rsintheta)^2 = 4$, or $r^2-4rcostheta = 0$. We're interested in the inside, which would be $r^2-4rcostheta le 0$. The other two bounding surfaces are $z=r$ as you found and $z=0$, for $0le zle r$.



          Second - no, $theta$ is not from zero to $2pi$. Look at the picture; the cylinder is entirely on one side of the $y$ axis. That's only half of the angle range around the origin, for $theta$ between $-fracpi2$ and $fracpi2$.



          Putting these pieces together, the integral becomes
          $$V = int_-pi/2^pi/2int_?^? (r-0)cdot r,dr,dtheta$$
          Yeah, we still need those $r$ bounds. That comes from the cylinder. Solving $r^2le 4rcostheta$ for $r$, we get (since $r$ is always nonnegative) that $0le rle 4costheta$. Now we finally have the integral completely set up:
          $$V = int_-pi/2^pi/2int_0^4costheta rcdot r,dr,dtheta$$
          Can you take it from there?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
            $endgroup$
            – krauser126
            Mar 27 at 20:16






          • 1




            $begingroup$
            The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
            $endgroup$
            – jmerry
            Mar 27 at 20:19


















          0












          $begingroup$

          Why use (shifted) radial coordinates when rectilinear are so easy?



          $$intlimits_x=0^4 intlimits_y = -sqrt4-(x-2)^2^sqrt4-(x-2)^2 intlimits_z=0^sqrtx^2 + y^2 1 dx dy dz = frac2569 neq 16 pi.$$



          enter image description here






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            What software did you use to get this graph?
            $endgroup$
            – user1952500
            Apr 8 at 2:57






          • 1




            $begingroup$
            Mathematica...
            $endgroup$
            – David G. Stork
            Apr 8 at 3:12











          Your Answer








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          2 Answers
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          2 Answers
          2






          active

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          active

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          active

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          0












          $begingroup$

          First - no, that's not the correct equation for the cylinder. There's no $z$ in the cylinder's equation; it should be $(rcostheta-2)^2+(rsintheta)^2 = 4$, or $r^2-4rcostheta = 0$. We're interested in the inside, which would be $r^2-4rcostheta le 0$. The other two bounding surfaces are $z=r$ as you found and $z=0$, for $0le zle r$.



          Second - no, $theta$ is not from zero to $2pi$. Look at the picture; the cylinder is entirely on one side of the $y$ axis. That's only half of the angle range around the origin, for $theta$ between $-fracpi2$ and $fracpi2$.



          Putting these pieces together, the integral becomes
          $$V = int_-pi/2^pi/2int_?^? (r-0)cdot r,dr,dtheta$$
          Yeah, we still need those $r$ bounds. That comes from the cylinder. Solving $r^2le 4rcostheta$ for $r$, we get (since $r$ is always nonnegative) that $0le rle 4costheta$. Now we finally have the integral completely set up:
          $$V = int_-pi/2^pi/2int_0^4costheta rcdot r,dr,dtheta$$
          Can you take it from there?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
            $endgroup$
            – krauser126
            Mar 27 at 20:16






          • 1




            $begingroup$
            The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
            $endgroup$
            – jmerry
            Mar 27 at 20:19















          0












          $begingroup$

          First - no, that's not the correct equation for the cylinder. There's no $z$ in the cylinder's equation; it should be $(rcostheta-2)^2+(rsintheta)^2 = 4$, or $r^2-4rcostheta = 0$. We're interested in the inside, which would be $r^2-4rcostheta le 0$. The other two bounding surfaces are $z=r$ as you found and $z=0$, for $0le zle r$.



          Second - no, $theta$ is not from zero to $2pi$. Look at the picture; the cylinder is entirely on one side of the $y$ axis. That's only half of the angle range around the origin, for $theta$ between $-fracpi2$ and $fracpi2$.



          Putting these pieces together, the integral becomes
          $$V = int_-pi/2^pi/2int_?^? (r-0)cdot r,dr,dtheta$$
          Yeah, we still need those $r$ bounds. That comes from the cylinder. Solving $r^2le 4rcostheta$ for $r$, we get (since $r$ is always nonnegative) that $0le rle 4costheta$. Now we finally have the integral completely set up:
          $$V = int_-pi/2^pi/2int_0^4costheta rcdot r,dr,dtheta$$
          Can you take it from there?






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
            $endgroup$
            – krauser126
            Mar 27 at 20:16






          • 1




            $begingroup$
            The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
            $endgroup$
            – jmerry
            Mar 27 at 20:19













          0












          0








          0





          $begingroup$

          First - no, that's not the correct equation for the cylinder. There's no $z$ in the cylinder's equation; it should be $(rcostheta-2)^2+(rsintheta)^2 = 4$, or $r^2-4rcostheta = 0$. We're interested in the inside, which would be $r^2-4rcostheta le 0$. The other two bounding surfaces are $z=r$ as you found and $z=0$, for $0le zle r$.



          Second - no, $theta$ is not from zero to $2pi$. Look at the picture; the cylinder is entirely on one side of the $y$ axis. That's only half of the angle range around the origin, for $theta$ between $-fracpi2$ and $fracpi2$.



          Putting these pieces together, the integral becomes
          $$V = int_-pi/2^pi/2int_?^? (r-0)cdot r,dr,dtheta$$
          Yeah, we still need those $r$ bounds. That comes from the cylinder. Solving $r^2le 4rcostheta$ for $r$, we get (since $r$ is always nonnegative) that $0le rle 4costheta$. Now we finally have the integral completely set up:
          $$V = int_-pi/2^pi/2int_0^4costheta rcdot r,dr,dtheta$$
          Can you take it from there?






          share|cite|improve this answer









          $endgroup$



          First - no, that's not the correct equation for the cylinder. There's no $z$ in the cylinder's equation; it should be $(rcostheta-2)^2+(rsintheta)^2 = 4$, or $r^2-4rcostheta = 0$. We're interested in the inside, which would be $r^2-4rcostheta le 0$. The other two bounding surfaces are $z=r$ as you found and $z=0$, for $0le zle r$.



          Second - no, $theta$ is not from zero to $2pi$. Look at the picture; the cylinder is entirely on one side of the $y$ axis. That's only half of the angle range around the origin, for $theta$ between $-fracpi2$ and $fracpi2$.



          Putting these pieces together, the integral becomes
          $$V = int_-pi/2^pi/2int_?^? (r-0)cdot r,dr,dtheta$$
          Yeah, we still need those $r$ bounds. That comes from the cylinder. Solving $r^2le 4rcostheta$ for $r$, we get (since $r$ is always nonnegative) that $0le rle 4costheta$. Now we finally have the integral completely set up:
          $$V = int_-pi/2^pi/2int_0^4costheta rcdot r,dr,dtheta$$
          Can you take it from there?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 27 at 19:51









          jmerryjmerry

          17k11633




          17k11633











          • $begingroup$
            Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
            $endgroup$
            – krauser126
            Mar 27 at 20:16






          • 1




            $begingroup$
            The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
            $endgroup$
            – jmerry
            Mar 27 at 20:19
















          • $begingroup$
            Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
            $endgroup$
            – krauser126
            Mar 27 at 20:16






          • 1




            $begingroup$
            The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
            $endgroup$
            – jmerry
            Mar 27 at 20:19















          $begingroup$
          Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
          $endgroup$
          – krauser126
          Mar 27 at 20:16




          $begingroup$
          Thanks, that makes sense. I solved the equation for the cylinder up to the point of [cylinder function] = 0 but I mistakenly substituted z instead for 0 after that. Regarding the r bounds, why isn't it just 0 to 2 or 2 to 4 since the base of the cylinder is a circle so r should go from center to outer edge of the base?
          $endgroup$
          – krauser126
          Mar 27 at 20:16




          1




          1




          $begingroup$
          The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
          $endgroup$
          – jmerry
          Mar 27 at 20:19




          $begingroup$
          The base of the cylinder is a circle, but it's not a circle centered at the origin of the coordinate system we're using. An equation "$r=c$" represents a circle centered at the origin, and the equation of a circle centered elsewhere looks different in polar coordinates. This is at least the second nicest case, a circle that passes through the origin.
          $endgroup$
          – jmerry
          Mar 27 at 20:19











          0












          $begingroup$

          Why use (shifted) radial coordinates when rectilinear are so easy?



          $$intlimits_x=0^4 intlimits_y = -sqrt4-(x-2)^2^sqrt4-(x-2)^2 intlimits_z=0^sqrtx^2 + y^2 1 dx dy dz = frac2569 neq 16 pi.$$



          enter image description here






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            What software did you use to get this graph?
            $endgroup$
            – user1952500
            Apr 8 at 2:57






          • 1




            $begingroup$
            Mathematica...
            $endgroup$
            – David G. Stork
            Apr 8 at 3:12















          0












          $begingroup$

          Why use (shifted) radial coordinates when rectilinear are so easy?



          $$intlimits_x=0^4 intlimits_y = -sqrt4-(x-2)^2^sqrt4-(x-2)^2 intlimits_z=0^sqrtx^2 + y^2 1 dx dy dz = frac2569 neq 16 pi.$$



          enter image description here






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            What software did you use to get this graph?
            $endgroup$
            – user1952500
            Apr 8 at 2:57






          • 1




            $begingroup$
            Mathematica...
            $endgroup$
            – David G. Stork
            Apr 8 at 3:12













          0












          0








          0





          $begingroup$

          Why use (shifted) radial coordinates when rectilinear are so easy?



          $$intlimits_x=0^4 intlimits_y = -sqrt4-(x-2)^2^sqrt4-(x-2)^2 intlimits_z=0^sqrtx^2 + y^2 1 dx dy dz = frac2569 neq 16 pi.$$



          enter image description here






          share|cite|improve this answer











          $endgroup$



          Why use (shifted) radial coordinates when rectilinear are so easy?



          $$intlimits_x=0^4 intlimits_y = -sqrt4-(x-2)^2^sqrt4-(x-2)^2 intlimits_z=0^sqrtx^2 + y^2 1 dx dy dz = frac2569 neq 16 pi.$$



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 27 at 23:39

























          answered Mar 27 at 20:11









          David G. StorkDavid G. Stork

          12.2k41836




          12.2k41836











          • $begingroup$
            What software did you use to get this graph?
            $endgroup$
            – user1952500
            Apr 8 at 2:57






          • 1




            $begingroup$
            Mathematica...
            $endgroup$
            – David G. Stork
            Apr 8 at 3:12
















          • $begingroup$
            What software did you use to get this graph?
            $endgroup$
            – user1952500
            Apr 8 at 2:57






          • 1




            $begingroup$
            Mathematica...
            $endgroup$
            – David G. Stork
            Apr 8 at 3:12















          $begingroup$
          What software did you use to get this graph?
          $endgroup$
          – user1952500
          Apr 8 at 2:57




          $begingroup$
          What software did you use to get this graph?
          $endgroup$
          – user1952500
          Apr 8 at 2:57




          1




          1




          $begingroup$
          Mathematica...
          $endgroup$
          – David G. Stork
          Apr 8 at 3:12




          $begingroup$
          Mathematica...
          $endgroup$
          – David G. Stork
          Apr 8 at 3:12

















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