Mixing
Parametric integration negative area?
$begingroup$
I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.
Since $A=int_a^b yfrac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.
Was my book wrong? Is my thinking correct?
Thank you in advance :)
integration definite-integrals coordinate-systems parametric area
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
$endgroup$
add a comment |
$begingroup$
I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.
Since $A=int_a^b yfrac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.
Was my book wrong? Is my thinking correct?
Thank you in advance :)
integration definite-integrals coordinate-systems parametric area
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
$endgroup$
add a comment |
$begingroup$
I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.
Since $A=int_a^b yfrac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.
Was my book wrong? Is my thinking correct?
Thank you in advance :)
integration definite-integrals coordinate-systems parametric area
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
$endgroup$
I know there is a question very similar to mine already here Why does using an integral to calculate an area sometimes return a negative value when using a parametric equation?
, but I am still a bit unsure after reading the answers as to when the integral gives you a positive area and when it gives you a negative area. My text book says that the area is positive if t traces out the curve clockwise and negative for anticlockwise, however I think there is more going on with regards to whether the curve is above or below the x axis.
Since $A=int_a^b yfrac{dx}{dt}dt$, my guess is that you get a positive answer if either the graph is above the x axis (y is positive) and the point is moving to the right (dx/dt positive) or the graph is below the x axis and the point is moving to the left. Then you get a negative answer if either the graph is below the x axis but is being traced out moving to the right, or the graph is above the x axis but is being traced out moving to the left.
Was my book wrong? Is my thinking correct?
Thank you in advance :)
integration definite-integrals coordinate-systems parametric area
integration definite-integrals coordinate-systems parametric area
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
edited Apr 13 '17 at 12:20
Community♦
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
asked Feb 19 '15 at 13:00
21joanna1221joanna12
1,0671616
1,0671616
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your thinking is correct, and your book is not wrong.
You say that that the integral $int_a^b y frac{dx}{dt},dt$ is positive when either $y>0$ and $frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
Your thinking is correct, but the book is wrong in general.
Assuming that $tin [a,b]$ with $a<b$, then considering $displaystyle int_a^b y frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $tin [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $displaystyle int_a^b y frac{dx}{dt}dt = dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=rcostheta, y=rsintheta$. We find that $rgeq0$ and $0leq theta < dfrac{pi}{2}$. It's quite easy to show that $dfrac{dtheta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=costheta, y=sintheta$ for $thetainleft[0,dfrac{pi}{2}right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-dfrac{pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y dfrac{dx}{dt}$, nothing more.
answered Jan 18 at 22:31
CatManDooCatManDoo
11
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
Your thinking is correct, and your book is not wrong.
You say that that the integral $int_a^b y frac{dx}{dt},dt$ is positive when either $y>0$ and $frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
Your thinking is correct, and your book is not wrong.
You say that that the integral $int_a^b y frac{dx}{dt},dt$ is positive when either $y>0$ and $frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
Your thinking is correct, and your book is not wrong.
You say that that the integral $int_a^b y frac{dx}{dt},dt$ is positive when either $y>0$ and $frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
Your thinking is correct, and your book is not wrong.
You say that that the integral $int_a^b y frac{dx}{dt},dt$ is positive when either $y>0$ and $frac{dx}{dt}>0$ (above the $x$-axis, moving left-to-right), or $y < 0$ and $frac{dx}{dt} < 0$ (below the $x$-axis, moving right-to-left).
In both of these cases, the particle moves clockwise with respect to the origin.
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
answered May 9 '16 at 13:31
Matthew LeingangMatthew Leingang
16.4k12244
16.4k12244
add a comment |
add a comment |
$begingroup$
Your thinking is correct, but the book is wrong in general.
Assuming that $tin [a,b]$ with $a<b$, then considering $displaystyle int_a^b y frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $tin [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $displaystyle int_a^b y frac{dx}{dt}dt = dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=rcostheta, y=rsintheta$. We find that $rgeq0$ and $0leq theta < dfrac{pi}{2}$. It's quite easy to show that $dfrac{dtheta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=costheta, y=sintheta$ for $thetainleft[0,dfrac{pi}{2}right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-dfrac{pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y dfrac{dx}{dt}$, nothing more.
answered Jan 18 at 22:31
CatManDooCatManDoo
11
$endgroup$
add a comment |
$begingroup$
Your thinking is correct, but the book is wrong in general.
Assuming that $tin [a,b]$ with $a<b$, then considering $displaystyle int_a^b y frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $tin [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $displaystyle int_a^b y frac{dx}{dt}dt = dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=rcostheta, y=rsintheta$. We find that $rgeq0$ and $0leq theta < dfrac{pi}{2}$. It's quite easy to show that $dfrac{dtheta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=costheta, y=sintheta$ for $thetainleft[0,dfrac{pi}{2}right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-dfrac{pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y dfrac{dx}{dt}$, nothing more.
answered Jan 18 at 22:31
CatManDooCatManDoo
11
$endgroup$
add a comment |
$begingroup$
Your thinking is correct, but the book is wrong in general.
Assuming that $tin [a,b]$ with $a<b$, then considering $displaystyle int_a^b y frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $tin [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $displaystyle int_a^b y frac{dx}{dt}dt = dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=rcostheta, y=rsintheta$. We find that $rgeq0$ and $0leq theta < dfrac{pi}{2}$. It's quite easy to show that $dfrac{dtheta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=costheta, y=sintheta$ for $thetainleft[0,dfrac{pi}{2}right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-dfrac{pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y dfrac{dx}{dt}$, nothing more.
answered Jan 18 at 22:31
CatManDooCatManDoo
11
$endgroup$
Your thinking is correct, but the book is wrong in general.
Assuming that $tin [a,b]$ with $a<b$, then considering $displaystyle int_a^b y frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $tin [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $displaystyle int_a^b y frac{dx}{dt}dt = dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=rcostheta, y=rsintheta$. We find that $rgeq0$ and $0leq theta < dfrac{pi}{2}$. It's quite easy to show that $dfrac{dtheta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=costheta, y=sintheta$ for $thetainleft[0,dfrac{pi}{2}right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-dfrac{pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y dfrac{dx}{dt}$, nothing more.
answered Jan 18 at 22:31
CatManDooCatManDoo
11
edited Jan 18 at 22:40
answered Jan 18 at 22:31
CatManDooCatManDoo
11
answered Jan 18 at 22:31
CatManDooCatManDoo
11
answered Jan 18 at 22:31
CatManDooCatManDoo
11
11
add a comment |
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