Mixing
Evaluate $int_alpha_0^alphafracsec^2alpha dalphaVertvecr-vecr_Q Vert$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?
For context I'm letting the original vectors at the end of the post, and the link to the physics problem.
beginalign
V &= D int_alpha_0^alphafracsec^2alpha dalphaVertvecr-vecr_Q Vert
tag1 \
Vertvecr-vecr_Q Vert &= sqrtr^2 + D^2 sec^2alpha - 2rD [sinthetasinphitanalpha + costheta]
tag2 \
alpha_0 &= arctan(-fracL2D) qquad alpha = arctan(fracL2D)
endalign
$$ vecr = rsinthetacosphi hati + rsinthetasinphi hatj + rcostheta hatk $$
$$ vecr_Q = Dtanalpha hatj + D hatk $$
Physics Post
definite-integrals trigonometric-integrals
edited Aug 1 at 0:45
amWhy
189k25219431
asked Jul 31 at 23:48
liuzp
85
add a comment |Â
up vote
0
down vote
favorite
I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?
For context I'm letting the original vectors at the end of the post, and the link to the physics problem.
beginalign
V &= D int_alpha_0^alphafracsec^2alpha dalphaVertvecr-vecr_Q Vert
tag1 \
Vertvecr-vecr_Q Vert &= sqrtr^2 + D^2 sec^2alpha - 2rD [sinthetasinphitanalpha + costheta]
tag2 \
alpha_0 &= arctan(-fracL2D) qquad alpha = arctan(fracL2D)
endalign
$$ vecr = rsinthetacosphi hati + rsinthetasinphi hatj + rcostheta hatk $$
$$ vecr_Q = Dtanalpha hatj + D hatk $$
Physics Post
definite-integrals trigonometric-integrals
edited Aug 1 at 0:45
amWhy
189k25219431
asked Jul 31 at 23:48
liuzp
85
2
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
then what is the variable of integration?
– Clayton
Aug 1 at 1:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?
For context I'm letting the original vectors at the end of the post, and the link to the physics problem.
beginalign
V &= D int_alpha_0^alphafracsec^2alpha dalphaVertvecr-vecr_Q Vert
tag1 \
Vertvecr-vecr_Q Vert &= sqrtr^2 + D^2 sec^2alpha - 2rD [sinthetasinphitanalpha + costheta]
tag2 \
alpha_0 &= arctan(-fracL2D) qquad alpha = arctan(fracL2D)
endalign
$$ vecr = rsinthetacosphi hati + rsinthetasinphi hatj + rcostheta hatk $$
$$ vecr_Q = Dtanalpha hatj + D hatk $$
Physics Post
definite-integrals trigonometric-integrals
edited Aug 1 at 0:45
amWhy
189k25219431
asked Jul 31 at 23:48
liuzp
85
I was solving a physics problem and I run in to an integral that I could not solve. Could you give me some tips to solve this kind of integral?
For context I'm letting the original vectors at the end of the post, and the link to the physics problem.
beginalign
V &= D int_alpha_0^alphafracsec^2alpha dalphaVertvecr-vecr_Q Vert
tag1 \
Vertvecr-vecr_Q Vert &= sqrtr^2 + D^2 sec^2alpha - 2rD [sinthetasinphitanalpha + costheta]
tag2 \
alpha_0 &= arctan(-fracL2D) qquad alpha = arctan(fracL2D)
endalign
$$ vecr = rsinthetacosphi hati + rsinthetasinphi hatj + rcostheta hatk $$
$$ vecr_Q = Dtanalpha hatj + D hatk $$
Physics Post
definite-integrals trigonometric-integrals
edited Aug 1 at 0:45
amWhy
189k25219431
asked Jul 31 at 23:48
liuzp
85
edited Aug 1 at 0:45
amWhy
189k25219431
edited Aug 1 at 0:45
amWhy
189k25219431
edited Aug 1 at 0:45
amWhy
189k25219431
189k25219431
asked Jul 31 at 23:48
liuzp
85
asked Jul 31 at 23:48
liuzp
85
asked Jul 31 at 23:48
liuzp
85
85
2
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
then what is the variable of integration?
– Clayton
Aug 1 at 1:12
add a comment |Â
2
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
then what is the variable of integration?
– Clayton
Aug 1 at 1:12
2
2
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
then what is the variable of integration?
– Clayton
Aug 1 at 1:12
then what is the variable of integration?
– Clayton
Aug 1 at 1:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
This integral as stated can be evaluated using the substitution $u=tanalpha$, which brings it to the form
$int_-delta^deltafracdusqrtau^2+bu+c=frac1sqrtasinh^-1Bigg(fracu+fracb2asqrtfrac4ac-b^24a^2Bigg)Bigg|^delta_-delta$
where $delta=fracL2D$, $a=D^2$ , $b=-2rDsinthetasinphi$ , $c=r^2+D^2-2rDcostheta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.
answered Aug 1 at 1:38
DinosaurEgg
1515
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
This integral as stated can be evaluated using the substitution $u=tanalpha$, which brings it to the form
$int_-delta^deltafracdusqrtau^2+bu+c=frac1sqrtasinh^-1Bigg(fracu+fracb2asqrtfrac4ac-b^24a^2Bigg)Bigg|^delta_-delta$
where $delta=fracL2D$, $a=D^2$ , $b=-2rDsinthetasinphi$ , $c=r^2+D^2-2rDcostheta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.
answered Aug 1 at 1:38
DinosaurEgg
1515
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
add a comment |Â
up vote
0
down vote
accepted
This integral as stated can be evaluated using the substitution $u=tanalpha$, which brings it to the form
$int_-delta^deltafracdusqrtau^2+bu+c=frac1sqrtasinh^-1Bigg(fracu+fracb2asqrtfrac4ac-b^24a^2Bigg)Bigg|^delta_-delta$
where $delta=fracL2D$, $a=D^2$ , $b=-2rDsinthetasinphi$ , $c=r^2+D^2-2rDcostheta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.
answered Aug 1 at 1:38
DinosaurEgg
1515
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This integral as stated can be evaluated using the substitution $u=tanalpha$, which brings it to the form
$int_-delta^deltafracdusqrtau^2+bu+c=frac1sqrtasinh^-1Bigg(fracu+fracb2asqrtfrac4ac-b^24a^2Bigg)Bigg|^delta_-delta$
where $delta=fracL2D$, $a=D^2$ , $b=-2rDsinthetasinphi$ , $c=r^2+D^2-2rDcostheta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.
answered Aug 1 at 1:38
DinosaurEgg
1515
This integral as stated can be evaluated using the substitution $u=tanalpha$, which brings it to the form
$int_-delta^deltafracdusqrtau^2+bu+c=frac1sqrtasinh^-1Bigg(fracu+fracb2asqrtfrac4ac-b^24a^2Bigg)Bigg|^delta_-delta$
where $delta=fracL2D$, $a=D^2$ , $b=-2rDsinthetasinphi$ , $c=r^2+D^2-2rDcostheta$.
I took a look at the physics however and it seems that you have not considered the fact that the image of the charged rod through the conducting sphere is not going to be a rod itself, but a circle because every point of the rod is mapped a different distance away from the center of the sphere. Taking this into account, I don't know how useful this calculation is going to be for you but I submit it anyway.
answered Aug 1 at 1:38
DinosaurEgg
1515
answered Aug 1 at 1:38
DinosaurEgg
1515
answered Aug 1 at 1:38
DinosaurEgg
1515
answered Aug 1 at 1:38
DinosaurEgg
1515
1515
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
add a comment |Â
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
Well, thanks, for the feedback, I'm going to revise the calculations, and, just one more thing since you stated it, the radius of the image would be constant? Learning a new integral is allways good, but eventualy I'm going to solve the problem.
– liuzp
Aug 1 at 2:41
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
[Here](en.wikipedia.org/wiki/… ) you can get started with a single charge, and then use superpositions to construct the whole rod out of infinitesimal charges at the appropriate distances from the origin. Every charge is mapped to a different distance than the charge next to it, and using the rule that the link provides,I found myself that the image is going to be bent into a circular arc of constant radius.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
You should fill in the details and be advised that the electric field of a uniformly charged circular arc doesn't have a nice solution in terms of simple elementary functions, so the integral you will do will be a tad more advanced.
– DinosaurEgg
Aug 1 at 2:55
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
And my professor saying that he solved in a single A4 paper. But, again thanks, you've helped more than you think.
– liuzp
Aug 1 at 3:00
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
Let us not doubt your professor's awesomeness :P Good luck!
– DinosaurEgg
Aug 1 at 4:02
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868590%2fevaluate-int-alpha-0-alpha-frac-sec2-alpha-d-alpha-vert-vecr%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
How can $alpha$ be a bound of the integral and the variable of integration?
– Clayton
Jul 31 at 23:51
I've put the values for the boundaries in the equation below the vectors modules
– liuzp
Jul 31 at 23:55
then what is the variable of integration?
– Clayton
Aug 1 at 1:12