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Center of mass of a planar lamina

1

$begingroup$

I have to find the center of mass of a planar lamina bounded by $x=0$, $y=1/2$, and $y=x$, with the density of the lamina being $x/(1-y^2)^1/2$.

I ended up drawing a picture, and it looks like a triangle. However, I set up my integral to find the mass, and I end up getting

$$int_0^1/2int_0^yfracxsqrt1-y^2,rmdx,rmdy$$

as my double integral to find the mass. I really don't like the look of the density function. I was thinking about converting everything to cylindrical coordinates, but I am not sure how to go about doing it. When I do come up with an integral in cylindrical coordinates, I think it looks even worse. Do any of you recommend converting to cylindrical coordinates?

In addition, I remember a previous calculation that I did where I found $(1-y^2)^1/2$ to be equal to $x$. I do not know where that identity comes from, and so I am afraid to substitute an $x$ for that part of the density.

calculus integration cylindrical-coordinates

share|cite|improve this question

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

$endgroup$

add a comment | 

1

$begingroup$

I have to find the center of mass of a planar lamina bounded by $x=0$, $y=1/2$, and $y=x$, with the density of the lamina being $x/(1-y^2)^1/2$.

I ended up drawing a picture, and it looks like a triangle. However, I set up my integral to find the mass, and I end up getting

$$int_0^1/2int_0^yfracxsqrt1-y^2,rmdx,rmdy$$

as my double integral to find the mass. I really don't like the look of the density function. I was thinking about converting everything to cylindrical coordinates, but I am not sure how to go about doing it. When I do come up with an integral in cylindrical coordinates, I think it looks even worse. Do any of you recommend converting to cylindrical coordinates?

In addition, I remember a previous calculation that I did where I found $(1-y^2)^1/2$ to be equal to $x$. I do not know where that identity comes from, and so I am afraid to substitute an $x$ for that part of the density.

calculus integration cylindrical-coordinates

share|cite|improve this question

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

$endgroup$

add a comment | 

$begingroup$

I have to find the center of mass of a planar lamina bounded by $x=0$, $y=1/2$, and $y=x$, with the density of the lamina being $x/(1-y^2)^1/2$.

I ended up drawing a picture, and it looks like a triangle. However, I set up my integral to find the mass, and I end up getting

$$int_0^1/2int_0^yfracxsqrt1-y^2,rmdx,rmdy$$

as my double integral to find the mass. I really don't like the look of the density function. I was thinking about converting everything to cylindrical coordinates, but I am not sure how to go about doing it. When I do come up with an integral in cylindrical coordinates, I think it looks even worse. Do any of you recommend converting to cylindrical coordinates?

In addition, I remember a previous calculation that I did where I found $(1-y^2)^1/2$ to be equal to $x$. I do not know where that identity comes from, and so I am afraid to substitute an $x$ for that part of the density.

calculus integration cylindrical-coordinates

share|cite|improve this question

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

$endgroup$

share|cite|improve this question

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

share|cite|improve this question

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

edited Mar 31 at 20:01

Robert Howard

2,3033935

edited Mar 31 at 20:01

Robert Howard

2,3033935

asked Mar 31 at 19:51

UchuukoUchuuko

317

asked Mar 31 at 19:51

UchuukoUchuuko

317

1 Answer
1

active

oldest

votes

2

$begingroup$

This integral can directly be performed
$$int_0^y fracxsqrt1-y^2dx=fracy^22sqrt1-y^2$$
$$int_0^frac12 fracy^22sqrt1-y^2dy=int_0^fracpi6 fracsin^2theta2sqrt1-sin^2theta costheta dtheta$$
$$=int_0^fracpi6 fracsin^2theta2dtheta$$
$$=int_0^fracpi6 frac1-cos2theta4dtheta$$
$$=[frac14(theta-frac12sin2theta)]_0^fracpi6$$
$$=fracpi24-fracsqrt316$$
Where I have used the substitution $y=sintheta$ in order to calculate the second integral.

share|cite|improve this answer

edited Mar 31 at 20:11

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see. Now I know that I don't have to convert at the beginning. Thank you so much.
  $endgroup$
  – Uchuuko
  Mar 31 at 20:20
  

  
  
  
  

add a comment | 

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2

$begingroup$

This integral can directly be performed
$$int_0^y fracxsqrt1-y^2dx=fracy^22sqrt1-y^2$$
$$int_0^frac12 fracy^22sqrt1-y^2dy=int_0^fracpi6 fracsin^2theta2sqrt1-sin^2theta costheta dtheta$$
$$=int_0^fracpi6 fracsin^2theta2dtheta$$
$$=int_0^fracpi6 frac1-cos2theta4dtheta$$
$$=[frac14(theta-frac12sin2theta)]_0^fracpi6$$
$$=fracpi24-fracsqrt316$$
Where I have used the substitution $y=sintheta$ in order to calculate the second integral.

share|cite|improve this answer

edited Mar 31 at 20:11

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see. Now I know that I don't have to convert at the beginning. Thank you so much.
  $endgroup$
  – Uchuuko
  Mar 31 at 20:20
  

  
  
  
  

add a comment | 

2

$begingroup$

This integral can directly be performed
$$int_0^y fracxsqrt1-y^2dx=fracy^22sqrt1-y^2$$
$$int_0^frac12 fracy^22sqrt1-y^2dy=int_0^fracpi6 fracsin^2theta2sqrt1-sin^2theta costheta dtheta$$
$$=int_0^fracpi6 fracsin^2theta2dtheta$$
$$=int_0^fracpi6 frac1-cos2theta4dtheta$$
$$=[frac14(theta-frac12sin2theta)]_0^fracpi6$$
$$=fracpi24-fracsqrt316$$
Where I have used the substitution $y=sintheta$ in order to calculate the second integral.

share|cite|improve this answer

edited Mar 31 at 20:11

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I see. Now I know that I don't have to convert at the beginning. Thank you so much.
  $endgroup$
  – Uchuuko
  Mar 31 at 20:20
  

  
  
  
  

add a comment | 

$begingroup$

This integral can directly be performed
$$int_0^y fracxsqrt1-y^2dx=fracy^22sqrt1-y^2$$
$$int_0^frac12 fracy^22sqrt1-y^2dy=int_0^fracpi6 fracsin^2theta2sqrt1-sin^2theta costheta dtheta$$
$$=int_0^fracpi6 fracsin^2theta2dtheta$$
$$=int_0^fracpi6 frac1-cos2theta4dtheta$$
$$=[frac14(theta-frac12sin2theta)]_0^fracpi6$$
$$=fracpi24-fracsqrt316$$
Where I have used the substitution $y=sintheta$ in order to calculate the second integral.

share|cite|improve this answer

edited Mar 31 at 20:11

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

$endgroup$

share|cite|improve this answer

edited Mar 31 at 20:11

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319

answered Mar 31 at 20:05

Peter ForemanPeter Foreman

7,2411319