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0

$begingroup$

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

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

$endgroup$

add a comment | 

0

$begingroup$

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

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

$endgroup$

add a comment | 

$begingroup$

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

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

$endgroup$

share|cite|improve this question

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

share|cite|improve this question

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

edited Mar 27 at 20:09

Gerschgorin

7810

edited Mar 27 at 20:09

Gerschgorin

7810

asked Mar 27 at 19:37

krauser126krauser126

636

asked Mar 27 at 19:37

krauser126krauser126

636

2 Answers
2

active

oldest

votes

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

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

0

$begingroup$

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

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

share|cite|improve this answer

edited Mar 27 at 23:39

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

$endgroup$

add a comment | 

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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

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

$endgroup$

share|cite|improve this answer

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

answered Mar 27 at 19:51

jmerryjmerry

16.9k11633

0

$begingroup$

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

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

share|cite|improve this answer

edited Mar 27 at 23:39

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

$endgroup$

add a comment | 

0

$begingroup$

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

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

share|cite|improve this answer

edited Mar 27 at 23:39

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

$endgroup$

add a comment | 

$begingroup$

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

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

share|cite|improve this answer

edited Mar 27 at 23:39

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

$endgroup$

share|cite|improve this answer

edited Mar 27 at 23:39

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533

answered Mar 27 at 20:11

David G. StorkDavid G. Stork

11.5k41533