Why does the sign of the second items is negative
$begingroup$
Consider
$$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
My question is why the sign of the second item is negative?
calculus integration
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
$endgroup$
add a comment |
$begingroup$
Consider
$$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
My question is why the sign of the second item is negative?
calculus integration
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
$endgroup$
add a comment |
$begingroup$
Consider
$$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
My question is why the sign of the second item is negative?
calculus integration
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
$endgroup$
Consider
$$int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy= int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
My question is why the sign of the second item is negative?
calculus integration
calculus integration
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
edited Dec 22 '18 at 7:47
Robert Z
96.1k1065136
96.1k1065136
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
asked Dec 22 '18 at 6:42
BloodpolyhydronBloodpolyhydron
41
41
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
$endgroup$
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$begingroup$
Here is the step-by-step solution:
$$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
&=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
Refer to the graph:
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
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add a comment |
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2 Answers
2
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2 Answers
2
active
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active
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active
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$begingroup$
In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
$endgroup$
In the integral on the $LHS$, in the region $pile xle2pi,-1le yle0,y$ ranges from $0tosin(x)$ for any value of $x$. In the corresponding term on the $RHS$, that is $displaystyleint_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx,y$ ranges from $-1to0$, which is the opposite direction. So we add a negative sign:$$-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx=int^{-1}_0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx$$
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
answered Dec 22 '18 at 7:48
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
add a comment |
$begingroup$
Here is the step-by-step solution:
$$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
&=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
Refer to the graph:
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
Here is the step-by-step solution:
$$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
&=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
Refer to the graph:
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
$endgroup$
add a comment |
$begingroup$
Here is the step-by-step solution:
$$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
&=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
Refer to the graph:
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
$endgroup$
Here is the step-by-step solution:
$$begin{align}int_0^{2pi}dxint_0^{sin(x)} f(x,y)dy&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy+int_{pi}^{2pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy=\
&=int_0^{pi}dxint_0^{color{green}{sin(x)}} f(x,y)dy-int_{pi}^{2pi}dxint_{color{blue}{sin(x)}}^0 f(x,y)dy=\
&=int_0^1dyint_{arcsin(y)}^{pi-arcsin(y)}f(x,y)dx-int_{-1}^0dyint_{pi-arcsin(y)}^{2pi+arcsin(y)}f(x,y)dx.end{align}$$
Refer to the graph:
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
answered Dec 22 '18 at 8:45
farruhotafarruhota
20.1k2738
20.1k2738
add a comment |
add a comment |
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