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Finding $iint_K y ,dxdy$, with $K$ bounded by the $x$-axis and one arch of the cycloid $x=R(t-sin t), y=R(1-cos t)$

Partial derivatives of $f(x_1, dots, x_n, g(x_1, dots, x_n))$How to obtain the gradient in polar coordinatesSurface integrals and surface areas of arbitrary parameter domainsTransforming the Bounds of an Integral for the Change of Variables TheoremFinding the volume bounded by a cylinder and a planearea of a plane inside an sphereApply Green's theorem for $iint_D (x+2y)dxdy$ on region bounded with $x = t- sin (t), y = 1 - cos (t)$.Evaluate the triple integral $iiint_E x,dV$ where $E$ is bounded by the paraboloid $x=4y^2+4z^2$ and the plane $x=4$.Show that $f_{xy}(0,0)=1$ and $f_{yx}(0,0)=-1$Determine the integral $f(x,y,z)=y$ bounded by the region $x=0,y=0,z=0$ and $2x+2y+z=4$

0

1

$begingroup$

I'm not sure how to approach this problem. I tried substituting the variables in the integrand, but I failed at expressing the bounds in terms of $x$ or $y$.

I thought: $0 le x le 2Rpi$ and $0 le y le dots$. (I'm not sure about the upper bound of $y$).

How can I solve this?

multivariable-calculus

share|cite|improve this question

asked Mar 11 at 18:57

ZacharyZachary

1799

$endgroup$

add a comment | 

0

1

$begingroup$

I'm not sure how to approach this problem. I tried substituting the variables in the integrand, but I failed at expressing the bounds in terms of $x$ or $y$.

I thought: $0 le x le 2Rpi$ and $0 le y le dots$. (I'm not sure about the upper bound of $y$).

How can I solve this?

multivariable-calculus

share|cite|improve this question

asked Mar 11 at 18:57

ZacharyZachary

1799

$endgroup$

add a comment | 

$begingroup$

I'm not sure how to approach this problem. I tried substituting the variables in the integrand, but I failed at expressing the bounds in terms of $x$ or $y$.

I thought: $0 le x le 2Rpi$ and $0 le y le dots$. (I'm not sure about the upper bound of $y$).

How can I solve this?

multivariable-calculus

share|cite|improve this question

asked Mar 11 at 18:57

ZacharyZachary

1799

$endgroup$

share|cite|improve this question

asked Mar 11 at 18:57

ZacharyZachary

1799

share|cite|improve this question

asked Mar 11 at 18:57

ZacharyZachary

1799

asked Mar 11 at 18:57

ZacharyZachary

1799

asked Mar 11 at 18:57

ZacharyZachary

1799

1 Answer
1

active

oldest

votes

1

$begingroup$

Since the function
$$psi:quad tmapsto x(t)=R(t-sin t)qquad(0leq tleq 2pi)$$
is monotonically increasing there is a function $f$ such that $$y(t)=fbigl(x(t)bigr)qquad(0leq tleq 2pi) ,$$
namely $f(x)=ybigl(psi^{-1}(x)bigr)$.
We can therefore assume that one arch $gamma$ of the cycloid is given by $$gamma:qquad y=f(x)quad (0leq xleq 2pi R) .$$ Then
$$J:=int_K y>{rm d}(x,y)=int_0^{2pi R}int_0^{f(x)} y >dy>dx={1over2}int_0^{2pi R}f^2(x)>dx .$$
We now substitute
$$x:=R(t-sin t)quad(0leq tleq 2pi) .$$
Then $dx=R(1-cos t)>dt$, and $f(x)=y=R(1-cos t)$,
so that
$$J={R^3over2}int_0^{2pi}(1-cos t)^2(1-cos t)>dt=ldotsquad.$$

share|cite|improve this answer

edited Mar 12 at 10:11

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. One more question regarding this solving technique: can I use this method for all curves that are expressed parameterically (f.ex. an astroid, $x=a cos^3 t, y = asin^3 t,, (0le t le pi/2)$)?
  $endgroup$
  – Zachary
  Mar 12 at 7:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Zachary: See my edit, and draw a figure!
  $endgroup$
  – Christian Blatter
  Mar 12 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function $psi: t to x(t) = a cos^3 t quad (0 le t le pi/2)$ is monotonous, therefore your method can be applied in this case too.
  $endgroup$
  – Zachary
  Mar 12 at 19:05
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Since the function
$$psi:quad tmapsto x(t)=R(t-sin t)qquad(0leq tleq 2pi)$$
is monotonically increasing there is a function $f$ such that $$y(t)=fbigl(x(t)bigr)qquad(0leq tleq 2pi) ,$$
namely $f(x)=ybigl(psi^{-1}(x)bigr)$.
We can therefore assume that one arch $gamma$ of the cycloid is given by $$gamma:qquad y=f(x)quad (0leq xleq 2pi R) .$$ Then
$$J:=int_K y>{rm d}(x,y)=int_0^{2pi R}int_0^{f(x)} y >dy>dx={1over2}int_0^{2pi R}f^2(x)>dx .$$
We now substitute
$$x:=R(t-sin t)quad(0leq tleq 2pi) .$$
Then $dx=R(1-cos t)>dt$, and $f(x)=y=R(1-cos t)$,
so that
$$J={R^3over2}int_0^{2pi}(1-cos t)^2(1-cos t)>dt=ldotsquad.$$

share|cite|improve this answer

edited Mar 12 at 10:11

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. One more question regarding this solving technique: can I use this method for all curves that are expressed parameterically (f.ex. an astroid, $x=a cos^3 t, y = asin^3 t,, (0le t le pi/2)$)?
  $endgroup$
  – Zachary
  Mar 12 at 7:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Zachary: See my edit, and draw a figure!
  $endgroup$
  – Christian Blatter
  Mar 12 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function $psi: t to x(t) = a cos^3 t quad (0 le t le pi/2)$ is monotonous, therefore your method can be applied in this case too.
  $endgroup$
  – Zachary
  Mar 12 at 19:05
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Since the function
$$psi:quad tmapsto x(t)=R(t-sin t)qquad(0leq tleq 2pi)$$
is monotonically increasing there is a function $f$ such that $$y(t)=fbigl(x(t)bigr)qquad(0leq tleq 2pi) ,$$
namely $f(x)=ybigl(psi^{-1}(x)bigr)$.
We can therefore assume that one arch $gamma$ of the cycloid is given by $$gamma:qquad y=f(x)quad (0leq xleq 2pi R) .$$ Then
$$J:=int_K y>{rm d}(x,y)=int_0^{2pi R}int_0^{f(x)} y >dy>dx={1over2}int_0^{2pi R}f^2(x)>dx .$$
We now substitute
$$x:=R(t-sin t)quad(0leq tleq 2pi) .$$
Then $dx=R(1-cos t)>dt$, and $f(x)=y=R(1-cos t)$,
so that
$$J={R^3over2}int_0^{2pi}(1-cos t)^2(1-cos t)>dt=ldotsquad.$$

share|cite|improve this answer

edited Mar 12 at 10:11

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much. One more question regarding this solving technique: can I use this method for all curves that are expressed parameterically (f.ex. an astroid, $x=a cos^3 t, y = asin^3 t,, (0le t le pi/2)$)?
  $endgroup$
  – Zachary
  Mar 12 at 7:02
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Zachary: See my edit, and draw a figure!
  $endgroup$
  – Christian Blatter
  Mar 12 at 10:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function $psi: t to x(t) = a cos^3 t quad (0 le t le pi/2)$ is monotonous, therefore your method can be applied in this case too.
  $endgroup$
  – Zachary
  Mar 12 at 19:05
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Since the function
$$psi:quad tmapsto x(t)=R(t-sin t)qquad(0leq tleq 2pi)$$
is monotonically increasing there is a function $f$ such that $$y(t)=fbigl(x(t)bigr)qquad(0leq tleq 2pi) ,$$
namely $f(x)=ybigl(psi^{-1}(x)bigr)$.
We can therefore assume that one arch $gamma$ of the cycloid is given by $$gamma:qquad y=f(x)quad (0leq xleq 2pi R) .$$ Then
$$J:=int_K y>{rm d}(x,y)=int_0^{2pi R}int_0^{f(x)} y >dy>dx={1over2}int_0^{2pi R}f^2(x)>dx .$$
We now substitute
$$x:=R(t-sin t)quad(0leq tleq 2pi) .$$
Then $dx=R(1-cos t)>dt$, and $f(x)=y=R(1-cos t)$,
so that
$$J={R^3over2}int_0^{2pi}(1-cos t)^2(1-cos t)>dt=ldotsquad.$$

share|cite|improve this answer

edited Mar 12 at 10:11

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

$endgroup$

share|cite|improve this answer

edited Mar 12 at 10:11

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327

answered Mar 11 at 19:38

Christian BlatterChristian Blatter

175k8115327