Mixing
Square root of a 2 variable quadratic - integration (2 different results - principal square root and factoring)
Clash Royale CLAN TAG#URR8PPP
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Consider the integral and the result according to wolfram alpha
$$ int_0^R sqrtr^2 + z^2 - 2zr;dr = fracr(r-2z)sqrt(r-z)^22(r-z)Bigg|_0^R \
=fracR^22 - zR
$$
where $z$ is a constant and $ 0 leq z < R$
Now lets factor the expression under the square root
$$ int_0^R sqrt(r - z)^2 ; dr$$
The square root only returns a magnitude according to convention (because we want it to be a function). So we have $|r - z|$ and our integral becomes
$$ int_0^z z - r ; dr + int_z^R r - z ; dr \
textor \
zr - fracr^22Bigg|_0^z + fracr^22 - zrBigg|_z^R \
fracR^22 - zR + z^2
$$
Why do I get 2 different results? Is the first method wrong and if so why? I must be integrating different functions if I'm getting different results. The 2nd method is the area under 2 line segments. The 1st method is the area under the slice of the surface $w = sqrtr^2 + z^2 -2zr$ where the slice is at some constant $z$. I'm confused what the math is saying in each situation. The 2 situations are different, but I don't know why they are different. It must somehow come down to the difference between $sqrtr^2 + z^2 - 2rz$ and $|r - z|$. Aren't $sqrt(x-a)^2$ and $|x-a|$ the same function? They can't be exactly the same because $int_0^10+adots ; dx$, depending on the integrand, gives different results?
calculus integration
asked Jul 20 at 14:53
DWade64
3891313
add a comment |Â
up vote
2
down vote
favorite
Consider the integral and the result according to wolfram alpha
$$ int_0^R sqrtr^2 + z^2 - 2zr;dr = fracr(r-2z)sqrt(r-z)^22(r-z)Bigg|_0^R \
=fracR^22 - zR
$$
where $z$ is a constant and $ 0 leq z < R$
Now lets factor the expression under the square root
$$ int_0^R sqrt(r - z)^2 ; dr$$
The square root only returns a magnitude according to convention (because we want it to be a function). So we have $|r - z|$ and our integral becomes
$$ int_0^z z - r ; dr + int_z^R r - z ; dr \
textor \
zr - fracr^22Bigg|_0^z + fracr^22 - zrBigg|_z^R \
fracR^22 - zR + z^2
$$
Why do I get 2 different results? Is the first method wrong and if so why? I must be integrating different functions if I'm getting different results. The 2nd method is the area under 2 line segments. The 1st method is the area under the slice of the surface $w = sqrtr^2 + z^2 -2zr$ where the slice is at some constant $z$. I'm confused what the math is saying in each situation. The 2 situations are different, but I don't know why they are different. It must somehow come down to the difference between $sqrtr^2 + z^2 - 2rz$ and $|r - z|$. Aren't $sqrt(x-a)^2$ and $|x-a|$ the same function? They can't be exactly the same because $int_0^10+adots ; dx$, depending on the integrand, gives different results?
calculus integration
asked Jul 20 at 14:53
DWade64
3891313
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the integral and the result according to wolfram alpha
$$ int_0^R sqrtr^2 + z^2 - 2zr;dr = fracr(r-2z)sqrt(r-z)^22(r-z)Bigg|_0^R \
=fracR^22 - zR
$$
where $z$ is a constant and $ 0 leq z < R$
Now lets factor the expression under the square root
$$ int_0^R sqrt(r - z)^2 ; dr$$
The square root only returns a magnitude according to convention (because we want it to be a function). So we have $|r - z|$ and our integral becomes
$$ int_0^z z - r ; dr + int_z^R r - z ; dr \
textor \
zr - fracr^22Bigg|_0^z + fracr^22 - zrBigg|_z^R \
fracR^22 - zR + z^2
$$
Why do I get 2 different results? Is the first method wrong and if so why? I must be integrating different functions if I'm getting different results. The 2nd method is the area under 2 line segments. The 1st method is the area under the slice of the surface $w = sqrtr^2 + z^2 -2zr$ where the slice is at some constant $z$. I'm confused what the math is saying in each situation. The 2 situations are different, but I don't know why they are different. It must somehow come down to the difference between $sqrtr^2 + z^2 - 2rz$ and $|r - z|$. Aren't $sqrt(x-a)^2$ and $|x-a|$ the same function? They can't be exactly the same because $int_0^10+adots ; dx$, depending on the integrand, gives different results?
calculus integration
asked Jul 20 at 14:53
DWade64
3891313
Consider the integral and the result according to wolfram alpha
$$ int_0^R sqrtr^2 + z^2 - 2zr;dr = fracr(r-2z)sqrt(r-z)^22(r-z)Bigg|_0^R \
=fracR^22 - zR
$$
where $z$ is a constant and $ 0 leq z < R$
Now lets factor the expression under the square root
$$ int_0^R sqrt(r - z)^2 ; dr$$
The square root only returns a magnitude according to convention (because we want it to be a function). So we have $|r - z|$ and our integral becomes
$$ int_0^z z - r ; dr + int_z^R r - z ; dr \
textor \
zr - fracr^22Bigg|_0^z + fracr^22 - zrBigg|_z^R \
fracR^22 - zR + z^2
$$
Why do I get 2 different results? Is the first method wrong and if so why? I must be integrating different functions if I'm getting different results. The 2nd method is the area under 2 line segments. The 1st method is the area under the slice of the surface $w = sqrtr^2 + z^2 -2zr$ where the slice is at some constant $z$. I'm confused what the math is saying in each situation. The 2 situations are different, but I don't know why they are different. It must somehow come down to the difference between $sqrtr^2 + z^2 - 2rz$ and $|r - z|$. Aren't $sqrt(x-a)^2$ and $|x-a|$ the same function? They can't be exactly the same because $int_0^10+adots ; dx$, depending on the integrand, gives different results?
calculus integration
asked Jul 20 at 14:53
DWade64
3891313
asked Jul 20 at 14:53
DWade64
3891313
asked Jul 20 at 14:53
DWade64
3891313
asked Jul 20 at 14:53
DWade64
3891313
3891313
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The problem is that the function
$$ r mapsto fracr(r-2z)lvert r-z rvert2 (r-z) = left(fracr^22 - r zright)operatornamesgn(r-z) $$
is discontinuous at $r=z$ . It is a valid antiderivative of $r mapsto lvert r-zrvert$ on the intervals $[0,z]$ and $[z,R]$ individually, but not on the whole interval $[0,R]$ , since it is of course not differentiable at $r=z$ as well.
If you split the interval of integration accordingly, you can still get the correct result:
$$ - left(fracr^22 - r zright) Bigg|_0^z + left(fracr^22 - r zright) Bigg|_z^R = fracR^22 - z R +z^2 , .$$
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The problem is that the function
$$ r mapsto fracr(r-2z)lvert r-z rvert2 (r-z) = left(fracr^22 - r zright)operatornamesgn(r-z) $$
is discontinuous at $r=z$ . It is a valid antiderivative of $r mapsto lvert r-zrvert$ on the intervals $[0,z]$ and $[z,R]$ individually, but not on the whole interval $[0,R]$ , since it is of course not differentiable at $r=z$ as well.
If you split the interval of integration accordingly, you can still get the correct result:
$$ - left(fracr^22 - r zright) Bigg|_0^z + left(fracr^22 - r zright) Bigg|_z^R = fracR^22 - z R +z^2 , .$$
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
add a comment |Â
up vote
2
down vote
accepted
The problem is that the function
$$ r mapsto fracr(r-2z)lvert r-z rvert2 (r-z) = left(fracr^22 - r zright)operatornamesgn(r-z) $$
is discontinuous at $r=z$ . It is a valid antiderivative of $r mapsto lvert r-zrvert$ on the intervals $[0,z]$ and $[z,R]$ individually, but not on the whole interval $[0,R]$ , since it is of course not differentiable at $r=z$ as well.
If you split the interval of integration accordingly, you can still get the correct result:
$$ - left(fracr^22 - r zright) Bigg|_0^z + left(fracr^22 - r zright) Bigg|_z^R = fracR^22 - z R +z^2 , .$$
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The problem is that the function
$$ r mapsto fracr(r-2z)lvert r-z rvert2 (r-z) = left(fracr^22 - r zright)operatornamesgn(r-z) $$
is discontinuous at $r=z$ . It is a valid antiderivative of $r mapsto lvert r-zrvert$ on the intervals $[0,z]$ and $[z,R]$ individually, but not on the whole interval $[0,R]$ , since it is of course not differentiable at $r=z$ as well.
If you split the interval of integration accordingly, you can still get the correct result:
$$ - left(fracr^22 - r zright) Bigg|_0^z + left(fracr^22 - r zright) Bigg|_z^R = fracR^22 - z R +z^2 , .$$
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
The problem is that the function
$$ r mapsto fracr(r-2z)lvert r-z rvert2 (r-z) = left(fracr^22 - r zright)operatornamesgn(r-z) $$
is discontinuous at $r=z$ . It is a valid antiderivative of $r mapsto lvert r-zrvert$ on the intervals $[0,z]$ and $[z,R]$ individually, but not on the whole interval $[0,R]$ , since it is of course not differentiable at $r=z$ as well.
If you split the interval of integration accordingly, you can still get the correct result:
$$ - left(fracr^22 - r zright) Bigg|_0^z + left(fracr^22 - r zright) Bigg|_z^R = fracR^22 - z R +z^2 , .$$
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
answered Jul 20 at 15:16
ComplexYetTrivial
2,607624
2,607624
add a comment |Â
add a comment |Â
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