Find the area enclosed by circles $r=3costheta$ and $r=3sintheta$

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Find the area enclosed by circles $r=3costheta$ and $r=3sintheta$

I know that the graph of $r=3sintheta$ intersects at $theta=0$ and at $theta=pi$ why aren't these two bounds used and why only $theta$ equaling $pi$? Also the graph $r=3costheta$ intersects the pole at $theta=frac3pi2,frac1pi2$. So why do they only use $fracpi2$?

calculus integration polar-coordinates

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asked Jul 24 at 0:10

Jinzu

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Find the area enclosed by circles $r=3costheta$ and $r=3sintheta$

I know that the graph of $r=3sintheta$ intersects at $theta=0$ and at $theta=pi$ why aren't these two bounds used and why only $theta$ equaling $pi$? Also the graph $r=3costheta$ intersects the pole at $theta=frac3pi2,frac1pi2$. So why do they only use $fracpi2$?

calculus integration polar-coordinates

share|cite|improve this question

asked Jul 24 at 0:10

Jinzu

328311

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Find the area enclosed by circles $r=3costheta$ and $r=3sintheta$

I know that the graph of $r=3sintheta$ intersects at $theta=0$ and at $theta=pi$ why aren't these two bounds used and why only $theta$ equaling $pi$? Also the graph $r=3costheta$ intersects the pole at $theta=frac3pi2,frac1pi2$. So why do they only use $fracpi2$?

calculus integration polar-coordinates

share|cite|improve this question

asked Jul 24 at 0:10

Jinzu

328311

Find the area enclosed by circles $r=3costheta$ and $r=3sintheta$

I know that the graph of $r=3sintheta$ intersects at $theta=0$ and at $theta=pi$ why aren't these two bounds used and why only $theta$ equaling $pi$? Also the graph $r=3costheta$ intersects the pole at $theta=frac3pi2,frac1pi2$. So why do they only use $fracpi2$?

calculus integration polar-coordinates

share|cite|improve this question

asked Jul 24 at 0:10

Jinzu

328311

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 0:10

Jinzu

328311

asked Jul 24 at 0:10

Jinzu

328311

asked Jul 24 at 0:10

Jinzu

328311

328311

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote



accepted

The two circles are intersecting at $theta=dfracpi4$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $theta=0mbox to theta=dfracpi4$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $theta=dfracpi4$ to $dfracpi2$

So, we get $$dfrac12int_0^dfracpi4(3sintheta)^2dtheta+dfrac12int_fracpi4^dfracpi2(3costheta)^2dtheta$$

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edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
  – Ted Shifrin
  Jul 24 at 0:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, but you still have a bunch of stuff to fix.
  – Ted Shifrin
  Jul 24 at 0:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
  – Ted Shifrin
  Jul 24 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
  – Ted Shifrin
  Jul 24 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TedShifrin Fixed
  – Key Flex
  Jul 24 at 0:57
  

  
  
  
  

add a comment | 

up vote
0
down vote

Answer to the question: "Why do they only use $pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $theta in left[0,;frac pi2right]$.

Consider now a "radar ray" of an angle $theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is
$$
beginaligned
J &=int_theta in [0,;pi/2]
int_rin[ 0,; 3min(costheta,; sintheta) ]r; dr; dtheta
\
&=
int_theta in [0,;pi/2]
left[frac r^22 right]_r=0^r=3min(costheta,;sintheta); dtheta
\
&=
2int_theta in [0,;pi/4]
frac 12cdot 9sin^2theta; dtheta
\
&qquadtext(use the substitution $y=pi/2-x$ for the
interval $left[fracpi4,;fracpi2right]$)
\
&=
frac 98(pi-2)
endaligned
$$
The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $frac 32$,




• and from it we remove half of the area of a square with side $frac 32$. I.e.
  $$
  2left( frac 14pileft(frac 32right)^2-frac 12left(frac 32right)^2 right) .
  $$

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answered Jul 24 at 1:53

dan_fulea

4,1271211

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up vote
0
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No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $pi/2$. So the area of that sector is $2.25pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25pi/4 -2.25/2 $ and the desired area is $5pi/4-2.25$

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answered Jul 24 at 2:28

Shrey Joshi

1389

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The two circles are intersecting at $theta=dfracpi4$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $theta=0mbox to theta=dfracpi4$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $theta=dfracpi4$ to $dfracpi2$

So, we get $$dfrac12int_0^dfracpi4(3sintheta)^2dtheta+dfrac12int_fracpi4^dfracpi2(3costheta)^2dtheta$$

share|cite|improve this answer

edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
  – Ted Shifrin
  Jul 24 at 0:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, but you still have a bunch of stuff to fix.
  – Ted Shifrin
  Jul 24 at 0:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
  – Ted Shifrin
  Jul 24 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
  – Ted Shifrin
  Jul 24 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TedShifrin Fixed
  – Key Flex
  Jul 24 at 0:57
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

The two circles are intersecting at $theta=dfracpi4$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $theta=0mbox to theta=dfracpi4$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $theta=dfracpi4$ to $dfracpi2$

So, we get $$dfrac12int_0^dfracpi4(3sintheta)^2dtheta+dfrac12int_fracpi4^dfracpi2(3costheta)^2dtheta$$

share|cite|improve this answer

edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
  – Ted Shifrin
  Jul 24 at 0:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, but you still have a bunch of stuff to fix.
  – Ted Shifrin
  Jul 24 at 0:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
  – Ted Shifrin
  Jul 24 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
  – Ted Shifrin
  Jul 24 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TedShifrin Fixed
  – Key Flex
  Jul 24 at 0:57
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The two circles are intersecting at $theta=dfracpi4$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $theta=0mbox to theta=dfracpi4$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $theta=dfracpi4$ to $dfracpi2$

So, we get $$dfrac12int_0^dfracpi4(3sintheta)^2dtheta+dfrac12int_fracpi4^dfracpi2(3costheta)^2dtheta$$

share|cite|improve this answer

edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

The two circles are intersecting at $theta=dfracpi4$ and $r=0$

The area region can be divided into two parts.

Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $theta=0mbox to theta=dfracpi4$

Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $theta=dfracpi4$ to $dfracpi2$

So, we get $$dfrac12int_0^dfracpi4(3sintheta)^2dtheta+dfrac12int_fracpi4^dfracpi2(3costheta)^2dtheta$$

share|cite|improve this answer

edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 0:57

edited Jul 24 at 0:57

edited Jul 24 at 0:57

answered Jul 24 at 0:20

Key Flex

4,233423

answered Jul 24 at 0:20

Key Flex

4,233423

answered Jul 24 at 0:20

Key Flex

4,233423

4,233423

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
  – Ted Shifrin
  Jul 24 at 0:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, but you still have a bunch of stuff to fix.
  – Ted Shifrin
  Jul 24 at 0:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
  – Ted Shifrin
  Jul 24 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
  – Ted Shifrin
  Jul 24 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TedShifrin Fixed
  – Key Flex
  Jul 24 at 0:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
  – Ted Shifrin
  Jul 24 at 0:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great, but you still have a bunch of stuff to fix.
  – Ted Shifrin
  Jul 24 at 0:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
  – Ted Shifrin
  Jul 24 at 0:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
  – Ted Shifrin
  Jul 24 at 0:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TedShifrin Fixed
  – Key Flex
  Jul 24 at 0:57
  

  
  
  
  

Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
– Ted Shifrin
Jul 24 at 0:28

Perhaps you should proofread? It's also worth commenting that the two circles intersect at $r=0$, even though the $theta$-values are different.
– Ted Shifrin
Jul 24 at 0:28

Great, but you still have a bunch of stuff to fix.
– Ted Shifrin
Jul 24 at 0:34

Great, but you still have a bunch of stuff to fix.
– Ted Shifrin
Jul 24 at 0:34

Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
– Ted Shifrin
Jul 24 at 0:36

Did you look at the sum of integrals you wrote down? It's a matter of having something correct, not a matter of adding.
– Ted Shifrin
Jul 24 at 0:36

Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
– Ted Shifrin
Jul 24 at 0:56

Agh. You need to be more self-critical. The integrand in the second integral is still wrong.
– Ted Shifrin
Jul 24 at 0:56

@TedShifrin Fixed
– Key Flex
Jul 24 at 0:57

@TedShifrin Fixed
– Key Flex
Jul 24 at 0:57

add a comment | 

up vote
0
down vote

Answer to the question: "Why do they only use $pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $theta in left[0,;frac pi2right]$.

Consider now a "radar ray" of an angle $theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is
$$
beginaligned
J &=int_theta in [0,;pi/2]
int_rin[ 0,; 3min(costheta,; sintheta) ]r; dr; dtheta
\
&=
int_theta in [0,;pi/2]
left[frac r^22 right]_r=0^r=3min(costheta,;sintheta); dtheta
\
&=
2int_theta in [0,;pi/4]
frac 12cdot 9sin^2theta; dtheta
\
&qquadtext(use the substitution $y=pi/2-x$ for the
interval $left[fracpi4,;fracpi2right]$)
\
&=
frac 98(pi-2)
endaligned
$$
The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $frac 32$,




• and from it we remove half of the area of a square with side $frac 32$. I.e.
  $$
  2left( frac 14pileft(frac 32right)^2-frac 12left(frac 32right)^2 right) .
  $$

share|cite|improve this answer

answered Jul 24 at 1:53

dan_fulea

4,1271211

add a comment | 

up vote
0
down vote

Answer to the question: "Why do they only use $pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $theta in left[0,;frac pi2right]$.

Consider now a "radar ray" of an angle $theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is
$$
beginaligned
J &=int_theta in [0,;pi/2]
int_rin[ 0,; 3min(costheta,; sintheta) ]r; dr; dtheta
\
&=
int_theta in [0,;pi/2]
left[frac r^22 right]_r=0^r=3min(costheta,;sintheta); dtheta
\
&=
2int_theta in [0,;pi/4]
frac 12cdot 9sin^2theta; dtheta
\
&qquadtext(use the substitution $y=pi/2-x$ for the
interval $left[fracpi4,;fracpi2right]$)
\
&=
frac 98(pi-2)
endaligned
$$
The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $frac 32$,




• and from it we remove half of the area of a square with side $frac 32$. I.e.
  $$
  2left( frac 14pileft(frac 32right)^2-frac 12left(frac 32right)^2 right) .
  $$

share|cite|improve this answer

answered Jul 24 at 1:53

dan_fulea

4,1271211

add a comment | 

up vote
0
down vote

up vote
0
down vote

Answer to the question: "Why do they only use $pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $theta in left[0,;frac pi2right]$.

Consider now a "radar ray" of an angle $theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is
$$
beginaligned
J &=int_theta in [0,;pi/2]
int_rin[ 0,; 3min(costheta,; sintheta) ]r; dr; dtheta
\
&=
int_theta in [0,;pi/2]
left[frac r^22 right]_r=0^r=3min(costheta,;sintheta); dtheta
\
&=
2int_theta in [0,;pi/4]
frac 12cdot 9sin^2theta; dtheta
\
&qquadtext(use the substitution $y=pi/2-x$ for the
interval $left[fracpi4,;fracpi2right]$)
\
&=
frac 98(pi-2)
endaligned
$$
The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $frac 32$,




• and from it we remove half of the area of a square with side $frac 32$. I.e.
  $$
  2left( frac 14pileft(frac 32right)^2-frac 12left(frac 32right)^2 right) .
  $$

share|cite|improve this answer

answered Jul 24 at 1:53

dan_fulea

4,1271211

Answer to the question: "Why do they only use $pi/2$... ?"

Note that the intersection region is locate in the first quadrant. This is parametrized by $theta in left[0,;frac pi2right]$.

Consider now a "radar ray" of an angle $theta$ in this interval. Then it first determines an $r$-interval in the intersection domain, and we know its limits. So the integral is
$$
beginaligned
J &=int_theta in [0,;pi/2]
int_rin[ 0,; 3min(costheta,; sintheta) ]r; dr; dtheta
\
&=
int_theta in [0,;pi/2]
left[frac r^22 right]_r=0^r=3min(costheta,;sintheta); dtheta
\
&=
2int_theta in [0,;pi/4]
frac 12cdot 9sin^2theta; dtheta
\
&qquadtext(use the substitution $y=pi/2-x$ for the
interval $left[fracpi4,;fracpi2right]$)
\
&=
frac 98(pi-2)
endaligned
$$
The result fits with the geometrical intuition, we compute the double of the following:

• the area of a quarter of a disk with radius $frac 32$,




• and from it we remove half of the area of a square with side $frac 32$. I.e.
  $$
  2left( frac 14pileft(frac 32right)^2-frac 12left(frac 32right)^2 right) .
  $$

share|cite|improve this answer

answered Jul 24 at 1:53

dan_fulea

4,1271211

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 1:53

dan_fulea

4,1271211

answered Jul 24 at 1:53

dan_fulea

4,1271211

answered Jul 24 at 1:53

dan_fulea

4,1271211

4,1271211

add a comment | 

add a comment | 

up vote
0
down vote

No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $pi/2$. So the area of that sector is $2.25pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25pi/4 -2.25/2 $ and the desired area is $5pi/4-2.25$

share|cite|improve this answer

answered Jul 24 at 2:28

Shrey Joshi

1389

add a comment | 

up vote
0
down vote

No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $pi/2$. So the area of that sector is $2.25pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25pi/4 -2.25/2 $ and the desired area is $5pi/4-2.25$

share|cite|improve this answer

answered Jul 24 at 2:28

Shrey Joshi

1389

add a comment | 

up vote
0
down vote

up vote
0
down vote

No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $pi/2$. So the area of that sector is $2.25pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25pi/4 -2.25/2 $ and the desired area is $5pi/4-2.25$

share|cite|improve this answer

answered Jul 24 at 2:28

Shrey Joshi

1389

No calculus:

Let the two intersection points of the circles be A and B do that A is (0, 0) and B is (1.5, 1.5). It's easy to see that the arc formed by the center of a circle and A and B has measure $pi/2$. So the area of that sector is $2.25pi/4$. The area of the triangle formed by A, B, and center of the circle is $2.25/2$. So the area of half the desired area is $2.25pi/4 -2.25/2 $ and the desired area is $5pi/4-2.25$

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answered Jul 24 at 2:28

Shrey Joshi

1389

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share|cite|improve this answer

answered Jul 24 at 2:28

Shrey Joshi

1389

answered Jul 24 at 2:28

Shrey Joshi

1389

answered Jul 24 at 2:28

Shrey Joshi

1389

1389

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