Determine an Integral..
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Let
$ B:= {(x,y,z) in mathbb{R}^3 | 1 leq z leq 3, x^2+ y^2 leq 16 } $
I want to determine
$ int_B frac{z}{cosh^2 sqrt{x^2+y^2}} d(x,y,z) $
I know it is a cylinder with radius 4.
I tried transforming with polarcoordinates-that was just not helpful. Can you give me some pretty hints how I finally this integral :-)? thanks in advance!
real-analysis integration
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constant94
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up vote
0
down vote
favorite
Let
$ B:= {(x,y,z) in mathbb{R}^3 | 1 leq z leq 3, x^2+ y^2 leq 16 } $
I want to determine
$ int_B frac{z}{cosh^2 sqrt{x^2+y^2}} d(x,y,z) $
I know it is a cylinder with radius 4.
I tried transforming with polarcoordinates-that was just not helpful. Can you give me some pretty hints how I finally this integral :-)? thanks in advance!
real-analysis integration
asked yesterday
constant94
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let
$ B:= {(x,y,z) in mathbb{R}^3 | 1 leq z leq 3, x^2+ y^2 leq 16 } $
I want to determine
$ int_B frac{z}{cosh^2 sqrt{x^2+y^2}} d(x,y,z) $
I know it is a cylinder with radius 4.
I tried transforming with polarcoordinates-that was just not helpful. Can you give me some pretty hints how I finally this integral :-)? thanks in advance!
real-analysis integration
asked yesterday
constant94
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let
$ B:= {(x,y,z) in mathbb{R}^3 | 1 leq z leq 3, x^2+ y^2 leq 16 } $
I want to determine
$ int_B frac{z}{cosh^2 sqrt{x^2+y^2}} d(x,y,z) $
I know it is a cylinder with radius 4.
I tried transforming with polarcoordinates-that was just not helpful. Can you give me some pretty hints how I finally this integral :-)? thanks in advance!
real-analysis integration
real-analysis integration
asked yesterday
constant94
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
constant94
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
constant94
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
constant94
132
asked yesterday
constant94
132
132
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
constant94 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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1 Answer
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1
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I think polar coordinates do work here:
$$int_B frac{z}{cosh^2sqrt{x^2+y^2}}dx,dy,dz=int_0^{2pi}!int_0^4!int_1^3frac{z r}{cosh^2 r}dz,dr,dtheta=2piint_1^3z,dzint_0^4frac{r}{cosh^2r},dr$$
so you just have to compute $intfrac{r}{cosh^2r},dr$. You can rewrite this as $intfrac{4r}{e^{-r}+e^r},dr=intfrac{4re^{2r}}{(1+e^{2r})^2},dr$ and try a substitution.
Edit: Actually the last integral is computed a lot more easily if you are familiar with hyperbolic functions:
$$intfrac{r}{cosh^2r},dr=int r,(tanh r)',dr=r,tanh r-inttanh r,dr=r,tanh r-ln(cosh r)$$
answered yesterday
ε-δ
23315
$8 pi tanh(4)$
– Alex Trounev
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I think polar coordinates do work here:
$$int_B frac{z}{cosh^2sqrt{x^2+y^2}}dx,dy,dz=int_0^{2pi}!int_0^4!int_1^3frac{z r}{cosh^2 r}dz,dr,dtheta=2piint_1^3z,dzint_0^4frac{r}{cosh^2r},dr$$
so you just have to compute $intfrac{r}{cosh^2r},dr$. You can rewrite this as $intfrac{4r}{e^{-r}+e^r},dr=intfrac{4re^{2r}}{(1+e^{2r})^2},dr$ and try a substitution.
Edit: Actually the last integral is computed a lot more easily if you are familiar with hyperbolic functions:
$$intfrac{r}{cosh^2r},dr=int r,(tanh r)',dr=r,tanh r-inttanh r,dr=r,tanh r-ln(cosh r)$$
answered yesterday
ε-δ
23315
$8 pi tanh(4)$
– Alex Trounev
yesterday
add a comment |
up vote
1
down vote
I think polar coordinates do work here:
$$int_B frac{z}{cosh^2sqrt{x^2+y^2}}dx,dy,dz=int_0^{2pi}!int_0^4!int_1^3frac{z r}{cosh^2 r}dz,dr,dtheta=2piint_1^3z,dzint_0^4frac{r}{cosh^2r},dr$$
so you just have to compute $intfrac{r}{cosh^2r},dr$. You can rewrite this as $intfrac{4r}{e^{-r}+e^r},dr=intfrac{4re^{2r}}{(1+e^{2r})^2},dr$ and try a substitution.
Edit: Actually the last integral is computed a lot more easily if you are familiar with hyperbolic functions:
$$intfrac{r}{cosh^2r},dr=int r,(tanh r)',dr=r,tanh r-inttanh r,dr=r,tanh r-ln(cosh r)$$
answered yesterday
ε-δ
23315
$8 pi tanh(4)$
– Alex Trounev
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
I think polar coordinates do work here:
$$int_B frac{z}{cosh^2sqrt{x^2+y^2}}dx,dy,dz=int_0^{2pi}!int_0^4!int_1^3frac{z r}{cosh^2 r}dz,dr,dtheta=2piint_1^3z,dzint_0^4frac{r}{cosh^2r},dr$$
so you just have to compute $intfrac{r}{cosh^2r},dr$. You can rewrite this as $intfrac{4r}{e^{-r}+e^r},dr=intfrac{4re^{2r}}{(1+e^{2r})^2},dr$ and try a substitution.
Edit: Actually the last integral is computed a lot more easily if you are familiar with hyperbolic functions:
$$intfrac{r}{cosh^2r},dr=int r,(tanh r)',dr=r,tanh r-inttanh r,dr=r,tanh r-ln(cosh r)$$
answered yesterday
ε-δ
23315
I think polar coordinates do work here:
$$int_B frac{z}{cosh^2sqrt{x^2+y^2}}dx,dy,dz=int_0^{2pi}!int_0^4!int_1^3frac{z r}{cosh^2 r}dz,dr,dtheta=2piint_1^3z,dzint_0^4frac{r}{cosh^2r},dr$$
so you just have to compute $intfrac{r}{cosh^2r},dr$. You can rewrite this as $intfrac{4r}{e^{-r}+e^r},dr=intfrac{4re^{2r}}{(1+e^{2r})^2},dr$ and try a substitution.
Edit: Actually the last integral is computed a lot more easily if you are familiar with hyperbolic functions:
$$intfrac{r}{cosh^2r},dr=int r,(tanh r)',dr=r,tanh r-inttanh r,dr=r,tanh r-ln(cosh r)$$
answered yesterday
ε-δ
23315
edited yesterday
answered yesterday
ε-δ
23315
answered yesterday
ε-δ
23315
answered yesterday
ε-δ
23315
23315
$8 pi tanh(4)$
– Alex Trounev
yesterday
add a comment |
$8 pi tanh(4)$
– Alex Trounev
yesterday
$8 pi tanh(4)$
– Alex Trounev
yesterday
$8 pi tanh(4)$
– Alex Trounev
yesterday
add a comment |
constant94 is a new contributor. Be nice, and check out our Code of Conduct.
constant94 is a new contributor. Be nice, and check out our Code of Conduct.
constant94 is a new contributor. Be nice, and check out our Code of Conduct.
constant94 is a new contributor. Be nice, and check out our Code of Conduct.
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