Find the area lying inside the cardioid $r=1+ costheta$ and outside the parabola $r(1+ costheta)=1$
$begingroup$
I need to find the area lying inside the cardioid $r=1+ costheta$ and outside the parabola $r(1+ costheta)=1$
ATTEMPT
First I found the intersection point of two curves which comes out to be $frac{-pi}{2}$ and $frac{pi}{2}$.
The integral setup will be $$int_{theta =frac{-pi}{2}}^{frac{pi}{2}}int_{r=frac{1}{1+costheta}}^{1+cos theta} dr dtheta$$, on integrating this I got the answer as $pi$ but answer was given to be $frac{3pi}{4}-frac 43$.
Can anybody check my integral setup?
multivariable-calculus polar-coordinates multiple-integral
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
$endgroup$
add a comment |
$begingroup$
I need to find the area lying inside the cardioid $r=1+ costheta$ and outside the parabola $r(1+ costheta)=1$
ATTEMPT
First I found the intersection point of two curves which comes out to be $frac{-pi}{2}$ and $frac{pi}{2}$.
The integral setup will be $$int_{theta =frac{-pi}{2}}^{frac{pi}{2}}int_{r=frac{1}{1+costheta}}^{1+cos theta} dr dtheta$$, on integrating this I got the answer as $pi$ but answer was given to be $frac{3pi}{4}-frac 43$.
Can anybody check my integral setup?
multivariable-calculus polar-coordinates multiple-integral
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
$endgroup$
$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20
add a comment |
$begingroup$
I need to find the area lying inside the cardioid $r=1+ costheta$ and outside the parabola $r(1+ costheta)=1$
ATTEMPT
First I found the intersection point of two curves which comes out to be $frac{-pi}{2}$ and $frac{pi}{2}$.
The integral setup will be $$int_{theta =frac{-pi}{2}}^{frac{pi}{2}}int_{r=frac{1}{1+costheta}}^{1+cos theta} dr dtheta$$, on integrating this I got the answer as $pi$ but answer was given to be $frac{3pi}{4}-frac 43$.
Can anybody check my integral setup?
multivariable-calculus polar-coordinates multiple-integral
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
$endgroup$
I need to find the area lying inside the cardioid $r=1+ costheta$ and outside the parabola $r(1+ costheta)=1$
ATTEMPT
First I found the intersection point of two curves which comes out to be $frac{-pi}{2}$ and $frac{pi}{2}$.
The integral setup will be $$int_{theta =frac{-pi}{2}}^{frac{pi}{2}}int_{r=frac{1}{1+costheta}}^{1+cos theta} dr dtheta$$, on integrating this I got the answer as $pi$ but answer was given to be $frac{3pi}{4}-frac 43$.
Can anybody check my integral setup?
multivariable-calculus polar-coordinates multiple-integral
multivariable-calculus polar-coordinates multiple-integral
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
edited Oct 3 '17 at 5:30
Martin Sleziak
44.9k10119273
44.9k10119273
asked Oct 1 '17 at 15:46
UserUser
1,88921229
asked Oct 1 '17 at 15:46
UserUser
1,88921229
asked Oct 1 '17 at 15:46
UserUser
1,88921229
1,88921229
$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20
add a comment |
$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20
$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20
add a comment |
1 Answer
1
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$begingroup$
You have done a simple mistake that is integration is done for ( r dr d theta) and not dr d there...
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
$endgroup$
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have done a simple mistake that is integration is done for ( r dr d theta) and not dr d there...
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
$endgroup$
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
add a comment |
$begingroup$
You have done a simple mistake that is integration is done for ( r dr d theta) and not dr d there...
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
$endgroup$
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
add a comment |
$begingroup$
You have done a simple mistake that is integration is done for ( r dr d theta) and not dr d there...
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
$endgroup$
You have done a simple mistake that is integration is done for ( r dr d theta) and not dr d there...
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
answered Nov 6 '17 at 19:11
Keshav BhattKeshav Bhatt
1
1
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
add a comment |
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
$begingroup$
Keshav, thanks for your answer. Please keep in mind that, when answering a question on this site, you should provide a more "full" answer. That is, tell the OP (the person who asked the question) how you arrived at your answer, and how your answer solves their problem.
$endgroup$
– fonini
Nov 6 '17 at 19:30
add a comment |
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$begingroup$
I think neither answer is right. I believe it is $3pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra
$endgroup$
– imranfat
Oct 1 '17 at 16:07
$begingroup$
Is the integral setup is fine?
$endgroup$
– User
Oct 1 '17 at 16:20