Determine an Integral..
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1
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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
asked Nov 20 at 22:10
constant94
625
add a comment |
up vote
1
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 Nov 20 at 22:10
constant94
625
add a comment |
up vote
1
down vote
favorite
up vote
1
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 Nov 20 at 22:10
constant94
625
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 Nov 20 at 22:10
constant94
625
asked Nov 20 at 22:10
constant94
625
asked Nov 20 at 22:10
constant94
625
asked Nov 20 at 22:10
constant94
625
asked Nov 20 at 22:10
constant94
625
625
add a comment |
add a comment |
1 Answer
1
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votes
up vote
2
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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 Nov 20 at 22:21
ε-δ
24315
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
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 Nov 20 at 22:21
ε-δ
24315
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
add a comment |
up vote
2
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 Nov 20 at 22:21
ε-δ
24315
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
add a comment |
up vote
2
down vote
up vote
2
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 Nov 20 at 22:21
ε-δ
24315
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 Nov 20 at 22:21
ε-δ
24315
edited Nov 20 at 23:33
answered Nov 20 at 22:21
ε-δ
24315
answered Nov 20 at 22:21
ε-δ
24315
answered Nov 20 at 22:21
ε-δ
24315
24315
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
add a comment |
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
$8 pi tanh(4)$
– Alex Trounev
Nov 20 at 22:34
add a comment |
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