How do you find the area of the ellipse $25x^2+4y^2=100$? [closed]

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How would you set up the integral using Jacobians and polar?

calculus jacobian

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edited Jul 27 at 23:46

Davide Morgante

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asked Jul 27 at 23:38

Hillary Oh

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closed as off-topic by amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u Jul 28 at 7:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why using integrals here, the area of an ellipse is defined as $A=pi ab$
  – Davide Morgante
  Jul 27 at 23:40
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

How would you set up the integral using Jacobians and polar?

calculus jacobian

share|cite|improve this question

edited Jul 27 at 23:46

Davide Morgante

1,724220

asked Jul 27 at 23:38

Hillary Oh

11

closed as off-topic by amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u Jul 28 at 7:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why using integrals here, the area of an ellipse is defined as $A=pi ab$
  – Davide Morgante
  Jul 27 at 23:40
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

How would you set up the integral using Jacobians and polar?

calculus jacobian

share|cite|improve this question

edited Jul 27 at 23:46

Davide Morgante

1,724220

asked Jul 27 at 23:38

Hillary Oh

11

How would you set up the integral using Jacobians and polar?

calculus jacobian

share|cite|improve this question

edited Jul 27 at 23:46

Davide Morgante

1,724220

asked Jul 27 at 23:38

Hillary Oh

11

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 23:46

Davide Morgante

1,724220

edited Jul 27 at 23:46

Davide Morgante

1,724220

edited Jul 27 at 23:46

Davide Morgante

1,724220

1,724220

asked Jul 27 at 23:38

Hillary Oh

11

asked Jul 27 at 23:38

Hillary Oh

11

asked Jul 27 at 23:38

Hillary Oh

11

11

closed as off-topic by amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u Jul 28 at 7:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u Jul 28 at 7:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, José Carlos Santos, Shailesh, Delta-u

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why using integrals here, the area of an ellipse is defined as $A=pi ab$
  – Davide Morgante
  Jul 27 at 23:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why using integrals here, the area of an ellipse is defined as $A=pi ab$
  – Davide Morgante
  Jul 27 at 23:40
  

  
  
  
  

1

1

Why using integrals here, the area of an ellipse is defined as $A=pi ab$
– Davide Morgante
Jul 27 at 23:40

Why using integrals here, the area of an ellipse is defined as $A=pi ab$
– Davide Morgante
Jul 27 at 23:40

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

By simply using the definition: roughly speaking

The area of a domain $D$ can be found by integration as $$A=iint_D dxdy$$

In our case the domain is defined as $$D=(x,y)inmathbbR^2:25x^2+4y^2=100$$ which being an ellipse, the area, can be calculated in elliptical coordinates. Before doing that we must go from your form to the general form for an ellipse, mainly $$fracx^2a^2+fracy^2b^2=1$$ which in your case, doing simple algebraic calculations (dividing both sides by $100$) is $$fracx^24+fracy^225=1$$ and we get the values $a=2$ and $b=5$. Now we use the elliptic transformations $$begincasesx=arhocos(theta) = 2rhocos(theta)\y=brhosin(theta)=5rhosin(theta)endcases$$ the Jacobean for the elliptic transformations is $$J = abrho drho dtheta;;;rhoin[0,1];;thetain[0,2pi]$$ The final integral then becomes $$A=int_0^2piint_0^1 10rho drho dtheta$$ which you can easily calculate

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

add a comment | 

up vote
2
down vote

The area of an ellipse is $pi ab$ and $a = 2, b = 5$. So, we should expect $10 pi$.

$x = frac 15 rcos theta\
y = frac 12 rsin theta$

The jacobian. -- $dx dy = frac 110 r dr dtheta$

$int_0^2piint_0^100 frac 110 r dr dtheta$

$10pi$

share|cite|improve this answer

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
  – abiessu
  Jul 27 at 23:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This seems wrong. Am I?
  – Randall
  Jul 27 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
  – Doug M
  Jul 27 at 23:59
  

  
  
  
  

add a comment | 

up vote
1
down vote

The equation of your ellipse is equivalent (when you divide both sides by 100) to $(frac x2)^2+(frac y5)^2=1$. That's the unit circle $x^2+y^2=1$ stretched by a factor 2 in the $x$-direction and by a factor $5$ in the $y$-direction. So the area is increased by a factor $2cdot5=10$. Since the unit circle bounds an area of $pi$, your ellipse bounds an area of $10pi$.

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

By simply using the definition: roughly speaking

The area of a domain $D$ can be found by integration as $$A=iint_D dxdy$$

In our case the domain is defined as $$D=(x,y)inmathbbR^2:25x^2+4y^2=100$$ which being an ellipse, the area, can be calculated in elliptical coordinates. Before doing that we must go from your form to the general form for an ellipse, mainly $$fracx^2a^2+fracy^2b^2=1$$ which in your case, doing simple algebraic calculations (dividing both sides by $100$) is $$fracx^24+fracy^225=1$$ and we get the values $a=2$ and $b=5$. Now we use the elliptic transformations $$begincasesx=arhocos(theta) = 2rhocos(theta)\y=brhosin(theta)=5rhosin(theta)endcases$$ the Jacobean for the elliptic transformations is $$J = abrho drho dtheta;;;rhoin[0,1];;thetain[0,2pi]$$ The final integral then becomes $$A=int_0^2piint_0^1 10rho drho dtheta$$ which you can easily calculate

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

add a comment | 

up vote
2
down vote

By simply using the definition: roughly speaking

The area of a domain $D$ can be found by integration as $$A=iint_D dxdy$$

In our case the domain is defined as $$D=(x,y)inmathbbR^2:25x^2+4y^2=100$$ which being an ellipse, the area, can be calculated in elliptical coordinates. Before doing that we must go from your form to the general form for an ellipse, mainly $$fracx^2a^2+fracy^2b^2=1$$ which in your case, doing simple algebraic calculations (dividing both sides by $100$) is $$fracx^24+fracy^225=1$$ and we get the values $a=2$ and $b=5$. Now we use the elliptic transformations $$begincasesx=arhocos(theta) = 2rhocos(theta)\y=brhosin(theta)=5rhosin(theta)endcases$$ the Jacobean for the elliptic transformations is $$J = abrho drho dtheta;;;rhoin[0,1];;thetain[0,2pi]$$ The final integral then becomes $$A=int_0^2piint_0^1 10rho drho dtheta$$ which you can easily calculate

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

add a comment | 

up vote
2
down vote

up vote
2
down vote

By simply using the definition: roughly speaking

The area of a domain $D$ can be found by integration as $$A=iint_D dxdy$$

In our case the domain is defined as $$D=(x,y)inmathbbR^2:25x^2+4y^2=100$$ which being an ellipse, the area, can be calculated in elliptical coordinates. Before doing that we must go from your form to the general form for an ellipse, mainly $$fracx^2a^2+fracy^2b^2=1$$ which in your case, doing simple algebraic calculations (dividing both sides by $100$) is $$fracx^24+fracy^225=1$$ and we get the values $a=2$ and $b=5$. Now we use the elliptic transformations $$begincasesx=arhocos(theta) = 2rhocos(theta)\y=brhosin(theta)=5rhosin(theta)endcases$$ the Jacobean for the elliptic transformations is $$J = abrho drho dtheta;;;rhoin[0,1];;thetain[0,2pi]$$ The final integral then becomes $$A=int_0^2piint_0^1 10rho drho dtheta$$ which you can easily calculate

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

By simply using the definition: roughly speaking

The area of a domain $D$ can be found by integration as $$A=iint_D dxdy$$

In our case the domain is defined as $$D=(x,y)inmathbbR^2:25x^2+4y^2=100$$ which being an ellipse, the area, can be calculated in elliptical coordinates. Before doing that we must go from your form to the general form for an ellipse, mainly $$fracx^2a^2+fracy^2b^2=1$$ which in your case, doing simple algebraic calculations (dividing both sides by $100$) is $$fracx^24+fracy^225=1$$ and we get the values $a=2$ and $b=5$. Now we use the elliptic transformations $$begincasesx=arhocos(theta) = 2rhocos(theta)\y=brhosin(theta)=5rhosin(theta)endcases$$ the Jacobean for the elliptic transformations is $$J = abrho drho dtheta;;;rhoin[0,1];;thetain[0,2pi]$$ The final integral then becomes $$A=int_0^2piint_0^1 10rho drho dtheta$$ which you can easily calculate

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 23:52

Davide Morgante

1,724220

answered Jul 27 at 23:52

Davide Morgante

1,724220

answered Jul 27 at 23:52

Davide Morgante

1,724220

1,724220

add a comment | 

add a comment | 

up vote
2
down vote

The area of an ellipse is $pi ab$ and $a = 2, b = 5$. So, we should expect $10 pi$.

$x = frac 15 rcos theta\
y = frac 12 rsin theta$

The jacobian. -- $dx dy = frac 110 r dr dtheta$

$int_0^2piint_0^100 frac 110 r dr dtheta$

$10pi$

share|cite|improve this answer

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
  – abiessu
  Jul 27 at 23:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This seems wrong. Am I?
  – Randall
  Jul 27 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
  – Doug M
  Jul 27 at 23:59
  

  
  
  
  

add a comment | 

up vote
2
down vote

The area of an ellipse is $pi ab$ and $a = 2, b = 5$. So, we should expect $10 pi$.

$x = frac 15 rcos theta\
y = frac 12 rsin theta$

The jacobian. -- $dx dy = frac 110 r dr dtheta$

$int_0^2piint_0^100 frac 110 r dr dtheta$

$10pi$

share|cite|improve this answer

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
  – abiessu
  Jul 27 at 23:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This seems wrong. Am I?
  – Randall
  Jul 27 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
  – Doug M
  Jul 27 at 23:59
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

The area of an ellipse is $pi ab$ and $a = 2, b = 5$. So, we should expect $10 pi$.

$x = frac 15 rcos theta\
y = frac 12 rsin theta$

The jacobian. -- $dx dy = frac 110 r dr dtheta$

$int_0^2piint_0^100 frac 110 r dr dtheta$

$10pi$

share|cite|improve this answer

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

The area of an ellipse is $pi ab$ and $a = 2, b = 5$. So, we should expect $10 pi$.

$x = frac 15 rcos theta\
y = frac 12 rsin theta$

The jacobian. -- $dx dy = frac 110 r dr dtheta$

$int_0^2piint_0^100 frac 110 r dr dtheta$

$10pi$

share|cite|improve this answer

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 23:58

edited Jul 27 at 23:58

edited Jul 27 at 23:58

answered Jul 27 at 23:43

Doug M

39.1k31749

answered Jul 27 at 23:43

Doug M

39.1k31749

answered Jul 27 at 23:43

Doug M

39.1k31749

39.1k31749

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
  – abiessu
  Jul 27 at 23:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This seems wrong. Am I?
  – Randall
  Jul 27 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
  – Doug M
  Jul 27 at 23:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
  – abiessu
  Jul 27 at 23:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This seems wrong. Am I?
  – Randall
  Jul 27 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
  – Doug M
  Jul 27 at 23:59
  

  
  
  
  

1

1

Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
– abiessu
Jul 27 at 23:56

Is it possible that the fractions for the values of $a,b$ are inverted? It seems like $y$ can take on values from $-5$ to $5$...
– abiessu
Jul 27 at 23:56

1

1

This seems wrong. Am I?
– Randall
Jul 27 at 23:57

This seems wrong. Am I?
– Randall
Jul 27 at 23:57

its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
– Doug M
Jul 27 at 23:59

its not that I have a,b inverted, I read the problem as $25x^2 + 4y^2 = 1$ so I was off by a factor of 100.
– Doug M
Jul 27 at 23:59

add a comment | 

up vote
1
down vote

The equation of your ellipse is equivalent (when you divide both sides by 100) to $(frac x2)^2+(frac y5)^2=1$. That's the unit circle $x^2+y^2=1$ stretched by a factor 2 in the $x$-direction and by a factor $5$ in the $y$-direction. So the area is increased by a factor $2cdot5=10$. Since the unit circle bounds an area of $pi$, your ellipse bounds an area of $10pi$.

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

add a comment | 

up vote
1
down vote

The equation of your ellipse is equivalent (when you divide both sides by 100) to $(frac x2)^2+(frac y5)^2=1$. That's the unit circle $x^2+y^2=1$ stretched by a factor 2 in the $x$-direction and by a factor $5$ in the $y$-direction. So the area is increased by a factor $2cdot5=10$. Since the unit circle bounds an area of $pi$, your ellipse bounds an area of $10pi$.

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

add a comment | 

up vote
1
down vote

up vote
1
down vote

The equation of your ellipse is equivalent (when you divide both sides by 100) to $(frac x2)^2+(frac y5)^2=1$. That's the unit circle $x^2+y^2=1$ stretched by a factor 2 in the $x$-direction and by a factor $5$ in the $y$-direction. So the area is increased by a factor $2cdot5=10$. Since the unit circle bounds an area of $pi$, your ellipse bounds an area of $10pi$.

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

The equation of your ellipse is equivalent (when you divide both sides by 100) to $(frac x2)^2+(frac y5)^2=1$. That's the unit circle $x^2+y^2=1$ stretched by a factor 2 in the $x$-direction and by a factor $5$ in the $y$-direction. So the area is increased by a factor $2cdot5=10$. Since the unit circle bounds an area of $pi$, your ellipse bounds an area of $10pi$.

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

answered Jul 28 at 0:32

Andreas Blass

47.5k348104

47.5k348104

add a comment | 

add a comment |