Mixing
Evaluating the triple integral $int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y}dzdydx$
$begingroup$
How do we evaluate the triple integral
$$int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y}dzdydx$$
I am totally struck at this easy looking problem. How do we change the orders so that the $y$ term in the denominator cancels out after integration? The second limits contain $x$ term. Any hints? Thanks beforehand.
real-analysis calculus multivariable-calculus
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
$endgroup$
add a comment |
$begingroup$
How do we evaluate the triple integral
$$int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y}dzdydx$$
I am totally struck at this easy looking problem. How do we change the orders so that the $y$ term in the denominator cancels out after integration? The second limits contain $x$ term. Any hints? Thanks beforehand.
real-analysis calculus multivariable-calculus
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
$endgroup$
1
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26
add a comment |
$begingroup$
How do we evaluate the triple integral
$$int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y}dzdydx$$
I am totally struck at this easy looking problem. How do we change the orders so that the $y$ term in the denominator cancels out after integration? The second limits contain $x$ term. Any hints? Thanks beforehand.
real-analysis calculus multivariable-calculus
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
$endgroup$
How do we evaluate the triple integral
$$int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y}dzdydx$$
I am totally struck at this easy looking problem. How do we change the orders so that the $y$ term in the denominator cancels out after integration? The second limits contain $x$ term. Any hints? Thanks beforehand.
real-analysis calculus multivariable-calculus
real-analysis calculus multivariable-calculus
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
asked Jan 28 at 12:07
vidyarthividyarthi
3,0731833
3,0731833
1
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26
add a comment |
1
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26
1
1
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all integrate in $z$:
$$
int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y},dzdydx=2int_0^{pi}int_x^{pi}frac{sin y}{y},dydx.
$$
For each fixed $xin[0,pi]$ $y$ moves between $x$ and $pi$. The region of integration is a triangle with vertices $(0,0)$, $(0,pi)$ and $(pi,pi)$. Changing the order of of integration, for each $yin[0,pi]$, $x$ moves between $0$ and $y$. The integral becomes
$$
2int_0^{pi}int_0^{y}frac{sin y}{y},dxdy,
$$
which is now easy to compute.
I strongly recommend to draw a picture of the region of integration to see what is going on.
edited Jan 28 at 14:12
Ankit Kumar
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
First of all integrate in $z$:
$$
int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y},dzdydx=2int_0^{pi}int_x^{pi}frac{sin y}{y},dydx.
$$
For each fixed $xin[0,pi]$ $y$ moves between $x$ and $pi$. The region of integration is a triangle with vertices $(0,0)$, $(0,pi)$ and $(pi,pi)$. Changing the order of of integration, for each $yin[0,pi]$, $x$ moves between $0$ and $y$. The integral becomes
$$
2int_0^{pi}int_0^{y}frac{sin y}{y},dxdy,
$$
which is now easy to compute.
I strongly recommend to draw a picture of the region of integration to see what is going on.
edited Jan 28 at 14:12
Ankit Kumar
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
add a comment |
$begingroup$
First of all integrate in $z$:
$$
int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y},dzdydx=2int_0^{pi}int_x^{pi}frac{sin y}{y},dydx.
$$
For each fixed $xin[0,pi]$ $y$ moves between $x$ and $pi$. The region of integration is a triangle with vertices $(0,0)$, $(0,pi)$ and $(pi,pi)$. Changing the order of of integration, for each $yin[0,pi]$, $x$ moves between $0$ and $y$. The integral becomes
$$
2int_0^{pi}int_0^{y}frac{sin y}{y},dxdy,
$$
which is now easy to compute.
I strongly recommend to draw a picture of the region of integration to see what is going on.
edited Jan 28 at 14:12
Ankit Kumar
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
add a comment |
$begingroup$
First of all integrate in $z$:
$$
int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y},dzdydx=2int_0^{pi}int_x^{pi}frac{sin y}{y},dydx.
$$
For each fixed $xin[0,pi]$ $y$ moves between $x$ and $pi$. The region of integration is a triangle with vertices $(0,0)$, $(0,pi)$ and $(pi,pi)$. Changing the order of of integration, for each $yin[0,pi]$, $x$ moves between $0$ and $y$. The integral becomes
$$
2int_0^{pi}int_0^{y}frac{sin y}{y},dxdy,
$$
which is now easy to compute.
I strongly recommend to draw a picture of the region of integration to see what is going on.
edited Jan 28 at 14:12
Ankit Kumar
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
First of all integrate in $z$:
$$
int_0^{pi}int_x^{pi}int_0^2frac{sin y}{y},dzdydx=2int_0^{pi}int_x^{pi}frac{sin y}{y},dydx.
$$
For each fixed $xin[0,pi]$ $y$ moves between $x$ and $pi$. The region of integration is a triangle with vertices $(0,0)$, $(0,pi)$ and $(pi,pi)$. Changing the order of of integration, for each $yin[0,pi]$, $x$ moves between $0$ and $y$. The integral becomes
$$
2int_0^{pi}int_0^{y}frac{sin y}{y},dxdy,
$$
which is now easy to compute.
I strongly recommend to draw a picture of the region of integration to see what is going on.
edited Jan 28 at 14:12
Ankit Kumar
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
edited Jan 28 at 14:12
Ankit Kumar
1,542221
edited Jan 28 at 14:12
Ankit Kumar
1,542221
edited Jan 28 at 14:12
Ankit Kumar
1,542221
1,542221
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 28 at 12:28
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
add a comment |
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
$begingroup$
thanks! I just got it. It was so easy
$endgroup$
– vidyarthi
Jan 28 at 12:29
add a comment |
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1
$begingroup$
The innermost integral is trivial.
$endgroup$
– random
Jan 28 at 12:25
$begingroup$
@random yes, but the very next integral does not have representation in elementary functions, if we proceed with the given order
$endgroup$
– vidyarthi
Jan 28 at 12:26