Mixing
general solution for double integral of square root of quadratic polynomials
$begingroup$
I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
int _0 ^1 int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx dy
$$
I have tried to solve
$$
int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx
$$
and this has a closed form of solution, but is very complicated that makes it nearlly impossible to get the integral further on y.
So I wonder if a closed form solution really exists, does anyone have some ideas?
integration multivariable-calculus
asked Jan 23 at 16:16
FengFeng
62
$endgroup$
add a comment |
$begingroup$
I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
int _0 ^1 int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx dy
$$
I have tried to solve
$$
int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx
$$
and this has a closed form of solution, but is very complicated that makes it nearlly impossible to get the integral further on y.
So I wonder if a closed form solution really exists, does anyone have some ideas?
integration multivariable-calculus
asked Jan 23 at 16:16
FengFeng
62
$endgroup$
1
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09
add a comment |
$begingroup$
I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
int _0 ^1 int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx dy
$$
I have tried to solve
$$
int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx
$$
and this has a closed form of solution, but is very complicated that makes it nearlly impossible to get the integral further on y.
So I wonder if a closed form solution really exists, does anyone have some ideas?
integration multivariable-calculus
asked Jan 23 at 16:16
FengFeng
62
$endgroup$
I wonder if there is a closed-form solution for double integral of square root of quadratic polynomials. Such as
$$
int _0 ^1 int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx dy
$$
I have tried to solve
$$
int _0 ^1 sqrt{a cdot x^2 + b cdot y^2 + c cdot x cdot y + d cdot x+ecdot y +1 } dx
$$
and this has a closed form of solution, but is very complicated that makes it nearlly impossible to get the integral further on y.
So I wonder if a closed form solution really exists, does anyone have some ideas?
integration multivariable-calculus
integration multivariable-calculus
asked Jan 23 at 16:16
FengFeng
62
asked Jan 23 at 16:16
FengFeng
62
edited Jan 26 at 19:51
Feng
asked Jan 23 at 16:16
FengFeng
62
asked Jan 23 at 16:16
FengFeng
62
asked Jan 23 at 16:16
FengFeng
62
62
1
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09
add a comment |
1
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09
1
1
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
MAPLE got:
$int _0^1int_0^1sqrt{acdot x^2+bcdot y^2+ccdot xcdot y+dcdot x+ecdot y +1 } dx dy$
$$=frac{1}{8}left(-4,ablnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,ablnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-4,aelnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,aelnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+{c}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{c}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+2,cdlnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-2,cdlnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,sqrt{a+b+c+d+e+1}{a}^{3/2}-4,sqrt{a+d+1}{a}^{3/2}+2,sqrt{a+b+c+d+e+1}sqrt{a}c-2,sqrt{b+e+1}sqrt{a}c+2,sqrt{a}d+2,sqrt{a+b+c+d+e+1}sqrt{a}d-2,sqrt{b+e+1}sqrt{a}d-2,sqrt{a+d+1}sqrt{a}dright){a}^{-3/2}$$
answered Jan 25 at 18:25
IV_IV_
1,511525
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
MAPLE got:
$int _0^1int_0^1sqrt{acdot x^2+bcdot y^2+ccdot xcdot y+dcdot x+ecdot y +1 } dx dy$
$$=frac{1}{8}left(-4,ablnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,ablnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-4,aelnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,aelnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+{c}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{c}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+2,cdlnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-2,cdlnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,sqrt{a+b+c+d+e+1}{a}^{3/2}-4,sqrt{a+d+1}{a}^{3/2}+2,sqrt{a+b+c+d+e+1}sqrt{a}c-2,sqrt{b+e+1}sqrt{a}c+2,sqrt{a}d+2,sqrt{a+b+c+d+e+1}sqrt{a}d-2,sqrt{b+e+1}sqrt{a}d-2,sqrt{a+d+1}sqrt{a}dright){a}^{-3/2}$$
answered Jan 25 at 18:25
IV_IV_
1,511525
$endgroup$
add a comment |
$begingroup$
MAPLE got:
$int _0^1int_0^1sqrt{acdot x^2+bcdot y^2+ccdot xcdot y+dcdot x+ecdot y +1 } dx dy$
$$=frac{1}{8}left(-4,ablnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,ablnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-4,aelnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,aelnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+{c}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{c}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+2,cdlnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-2,cdlnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,sqrt{a+b+c+d+e+1}{a}^{3/2}-4,sqrt{a+d+1}{a}^{3/2}+2,sqrt{a+b+c+d+e+1}sqrt{a}c-2,sqrt{b+e+1}sqrt{a}c+2,sqrt{a}d+2,sqrt{a+b+c+d+e+1}sqrt{a}d-2,sqrt{b+e+1}sqrt{a}d-2,sqrt{a+d+1}sqrt{a}dright){a}^{-3/2}$$
answered Jan 25 at 18:25
IV_IV_
1,511525
$endgroup$
add a comment |
$begingroup$
MAPLE got:
$int _0^1int_0^1sqrt{acdot x^2+bcdot y^2+ccdot xcdot y+dcdot x+ecdot y +1 } dx dy$
$$=frac{1}{8}left(-4,ablnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,ablnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-4,aelnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,aelnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+{c}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{c}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+2,cdlnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-2,cdlnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,sqrt{a+b+c+d+e+1}{a}^{3/2}-4,sqrt{a+d+1}{a}^{3/2}+2,sqrt{a+b+c+d+e+1}sqrt{a}c-2,sqrt{b+e+1}sqrt{a}c+2,sqrt{a}d+2,sqrt{a+b+c+d+e+1}sqrt{a}d-2,sqrt{b+e+1}sqrt{a}d-2,sqrt{a+d+1}sqrt{a}dright){a}^{-3/2}$$
answered Jan 25 at 18:25
IV_IV_
1,511525
$endgroup$
MAPLE got:
$int _0^1int_0^1sqrt{acdot x^2+bcdot y^2+ccdot xcdot y+dcdot x+ecdot y +1 } dx dy$
$$=frac{1}{8}left(-4,ablnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,ablnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-4,aelnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,aelnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+{c}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{c}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+2,cdlnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-2,cdlnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)+{d}^{2}lnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)-{d}^{2}lnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{d+2,sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{2,a+d+2,sqrt{a+d+1}sqrt{a}}{sqrt{a}}}right)-4,alnleft({frac{c+d+2,sqrt{b+e+1}sqrt{a}}{sqrt{a}}}right)+4,alnleft({frac{2,a+c+d+2,sqrt{a+b+c+d+e+1}sqrt{a}}{sqrt{a}}}right)+4,sqrt{a+b+c+d+e+1}{a}^{3/2}-4,sqrt{a+d+1}{a}^{3/2}+2,sqrt{a+b+c+d+e+1}sqrt{a}c-2,sqrt{b+e+1}sqrt{a}c+2,sqrt{a}d+2,sqrt{a+b+c+d+e+1}sqrt{a}d-2,sqrt{b+e+1}sqrt{a}d-2,sqrt{a+d+1}sqrt{a}dright){a}^{-3/2}$$
answered Jan 25 at 18:25
IV_IV_
1,511525
edited Jan 27 at 13:43
answered Jan 25 at 18:25
IV_IV_
1,511525
answered Jan 25 at 18:25
IV_IV_
1,511525
answered Jan 25 at 18:25
IV_IV_
1,511525
1,511525
add a comment |
add a comment |
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1
$begingroup$
Would it help to rotate in the $(x,y)$-plane to get rid of the $xy$-term, namely $cxy$? Then couldn’t you complete the square (shift the origin) to get your integrand to be something like $(a'xi^2+b'eta^2)^{1/2}$ ?
$endgroup$
– Lubin
Jan 23 at 16:22
$begingroup$
@Lubin thank you very much for your comment! But this only transforms the complexity from the polynomial to the domain and therefore does not simplify the final work.
$endgroup$
– Feng
Jan 25 at 17:09