Mixing
Double integral $x^2(x^2+y^2+1)^{3/2}$
$begingroup$
How to evaluate $int_{-1}^1 int_{-1}^1 x^2(x^2+y^2+1)^{3/2} dx dy$ ? I tried substituting $x^2+y^2=tan^2(u)$ but it becomes more complicated. Using Wolfram Alpha Widget the answer is $3.64092$. Any idea? Thanks.
calculus integration multivariable-calculus
asked Jan 6 at 4:10
TEDTED
34
$endgroup$
|
show 1 more comment
$begingroup$
How to evaluate $int_{-1}^1 int_{-1}^1 x^2(x^2+y^2+1)^{3/2} dx dy$ ? I tried substituting $x^2+y^2=tan^2(u)$ but it becomes more complicated. Using Wolfram Alpha Widget the answer is $3.64092$. Any idea? Thanks.
calculus integration multivariable-calculus
asked Jan 6 at 4:10
TEDTED
34
$endgroup$
$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
1
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
2
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
1
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01
|
show 1 more comment
$begingroup$
How to evaluate $int_{-1}^1 int_{-1}^1 x^2(x^2+y^2+1)^{3/2} dx dy$ ? I tried substituting $x^2+y^2=tan^2(u)$ but it becomes more complicated. Using Wolfram Alpha Widget the answer is $3.64092$. Any idea? Thanks.
calculus integration multivariable-calculus
asked Jan 6 at 4:10
TEDTED
34
$endgroup$
How to evaluate $int_{-1}^1 int_{-1}^1 x^2(x^2+y^2+1)^{3/2} dx dy$ ? I tried substituting $x^2+y^2=tan^2(u)$ but it becomes more complicated. Using Wolfram Alpha Widget the answer is $3.64092$. Any idea? Thanks.
calculus integration multivariable-calculus
calculus integration multivariable-calculus
asked Jan 6 at 4:10
TEDTED
34
asked Jan 6 at 4:10
TEDTED
34
edited Jan 6 at 4:23
TED
asked Jan 6 at 4:10
TEDTED
34
asked Jan 6 at 4:10
TEDTED
34
asked Jan 6 at 4:10
TEDTED
34
34
$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
1
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
2
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
1
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01
|
show 1 more comment
$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
1
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
2
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
1
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01
$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
1
1
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
2
2
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
1
1
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01
|
show 1 more comment
1 Answer
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$begingroup$
The symmetry of the integrand implies $$I = int_{y=-1}^1 int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} , dx , dy = int_{x=-1}^1 int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} , dy , dx.$$ Thus $$2I = int_{y=-1}^1 int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} , dx , dy.$$ Now perform the coordinate transformation $$(x,y) = (r cos theta, r sin theta)$$ to obtain $$2I = 8 int_{theta = 0}^{pi/4} int_{r = 0}^{sec theta} r^2 (r^2 + 1)^{3/2} r , dr , dtheta.$$ Since
$$begin{align*} int r^3 (r^2 + 1)^{3/2} , dr
&= frac{1}{2} int r^2 (r^2 + 1)^{3/2} 2r , dr \
&= frac{1}{2} int (u-1)u^{3/2} , du \
&= frac{1}{2} left( frac{2}{7} u^{7/2} - frac{2}{5} u^{5/2} right) + C \
&= frac{(r^2+1)^{7/2}}{7} - frac{(r^2+1)^{5/2}}{5} + C,
end{align*}$$
we obtain
$$
I = 4 int_{theta=0}^{pi/4} frac{(sec^2 theta + 1)^{7/2} - 1}{7} - frac{(sec^2 theta + 1)^{5/2} - 1}{5} , dtheta.$$
At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = frac{2}{105}left(96 sqrt{3} + pi + 33 coth^{-1} sqrt{3} right) approx 3.640919532942123602084014417112851989279ldots.$$
answered Jan 6 at 5:09
heropupheropup
63k66199
$endgroup$
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The symmetry of the integrand implies $$I = int_{y=-1}^1 int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} , dx , dy = int_{x=-1}^1 int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} , dy , dx.$$ Thus $$2I = int_{y=-1}^1 int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} , dx , dy.$$ Now perform the coordinate transformation $$(x,y) = (r cos theta, r sin theta)$$ to obtain $$2I = 8 int_{theta = 0}^{pi/4} int_{r = 0}^{sec theta} r^2 (r^2 + 1)^{3/2} r , dr , dtheta.$$ Since
$$begin{align*} int r^3 (r^2 + 1)^{3/2} , dr
&= frac{1}{2} int r^2 (r^2 + 1)^{3/2} 2r , dr \
&= frac{1}{2} int (u-1)u^{3/2} , du \
&= frac{1}{2} left( frac{2}{7} u^{7/2} - frac{2}{5} u^{5/2} right) + C \
&= frac{(r^2+1)^{7/2}}{7} - frac{(r^2+1)^{5/2}}{5} + C,
end{align*}$$
we obtain
$$
I = 4 int_{theta=0}^{pi/4} frac{(sec^2 theta + 1)^{7/2} - 1}{7} - frac{(sec^2 theta + 1)^{5/2} - 1}{5} , dtheta.$$
At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = frac{2}{105}left(96 sqrt{3} + pi + 33 coth^{-1} sqrt{3} right) approx 3.640919532942123602084014417112851989279ldots.$$
answered Jan 6 at 5:09
heropupheropup
63k66199
$endgroup$
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
add a comment |
$begingroup$
The symmetry of the integrand implies $$I = int_{y=-1}^1 int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} , dx , dy = int_{x=-1}^1 int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} , dy , dx.$$ Thus $$2I = int_{y=-1}^1 int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} , dx , dy.$$ Now perform the coordinate transformation $$(x,y) = (r cos theta, r sin theta)$$ to obtain $$2I = 8 int_{theta = 0}^{pi/4} int_{r = 0}^{sec theta} r^2 (r^2 + 1)^{3/2} r , dr , dtheta.$$ Since
$$begin{align*} int r^3 (r^2 + 1)^{3/2} , dr
&= frac{1}{2} int r^2 (r^2 + 1)^{3/2} 2r , dr \
&= frac{1}{2} int (u-1)u^{3/2} , du \
&= frac{1}{2} left( frac{2}{7} u^{7/2} - frac{2}{5} u^{5/2} right) + C \
&= frac{(r^2+1)^{7/2}}{7} - frac{(r^2+1)^{5/2}}{5} + C,
end{align*}$$
we obtain
$$
I = 4 int_{theta=0}^{pi/4} frac{(sec^2 theta + 1)^{7/2} - 1}{7} - frac{(sec^2 theta + 1)^{5/2} - 1}{5} , dtheta.$$
At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = frac{2}{105}left(96 sqrt{3} + pi + 33 coth^{-1} sqrt{3} right) approx 3.640919532942123602084014417112851989279ldots.$$
answered Jan 6 at 5:09
heropupheropup
63k66199
$endgroup$
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
add a comment |
$begingroup$
The symmetry of the integrand implies $$I = int_{y=-1}^1 int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} , dx , dy = int_{x=-1}^1 int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} , dy , dx.$$ Thus $$2I = int_{y=-1}^1 int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} , dx , dy.$$ Now perform the coordinate transformation $$(x,y) = (r cos theta, r sin theta)$$ to obtain $$2I = 8 int_{theta = 0}^{pi/4} int_{r = 0}^{sec theta} r^2 (r^2 + 1)^{3/2} r , dr , dtheta.$$ Since
$$begin{align*} int r^3 (r^2 + 1)^{3/2} , dr
&= frac{1}{2} int r^2 (r^2 + 1)^{3/2} 2r , dr \
&= frac{1}{2} int (u-1)u^{3/2} , du \
&= frac{1}{2} left( frac{2}{7} u^{7/2} - frac{2}{5} u^{5/2} right) + C \
&= frac{(r^2+1)^{7/2}}{7} - frac{(r^2+1)^{5/2}}{5} + C,
end{align*}$$
we obtain
$$
I = 4 int_{theta=0}^{pi/4} frac{(sec^2 theta + 1)^{7/2} - 1}{7} - frac{(sec^2 theta + 1)^{5/2} - 1}{5} , dtheta.$$
At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = frac{2}{105}left(96 sqrt{3} + pi + 33 coth^{-1} sqrt{3} right) approx 3.640919532942123602084014417112851989279ldots.$$
answered Jan 6 at 5:09
heropupheropup
63k66199
$endgroup$
The symmetry of the integrand implies $$I = int_{y=-1}^1 int_{x=-1}^1 x^2 (x^2 + y^2 + 1)^{3/2} , dx , dy = int_{x=-1}^1 int_{y=-1}^1 y^2 (x^2 + y^2 + 1)^{3/2} , dy , dx.$$ Thus $$2I = int_{y=-1}^1 int_{x=-1}^1 (x^2 + y^2)(x^2 + y^2 + 1)^{3/2} , dx , dy.$$ Now perform the coordinate transformation $$(x,y) = (r cos theta, r sin theta)$$ to obtain $$2I = 8 int_{theta = 0}^{pi/4} int_{r = 0}^{sec theta} r^2 (r^2 + 1)^{3/2} r , dr , dtheta.$$ Since
$$begin{align*} int r^3 (r^2 + 1)^{3/2} , dr
&= frac{1}{2} int r^2 (r^2 + 1)^{3/2} 2r , dr \
&= frac{1}{2} int (u-1)u^{3/2} , du \
&= frac{1}{2} left( frac{2}{7} u^{7/2} - frac{2}{5} u^{5/2} right) + C \
&= frac{(r^2+1)^{7/2}}{7} - frac{(r^2+1)^{5/2}}{5} + C,
end{align*}$$
we obtain
$$
I = 4 int_{theta=0}^{pi/4} frac{(sec^2 theta + 1)^{7/2} - 1}{7} - frac{(sec^2 theta + 1)^{5/2} - 1}{5} , dtheta.$$
At this point, I don't have a simple way to evaluate this integral; however, the value provided is $$I = frac{2}{105}left(96 sqrt{3} + pi + 33 coth^{-1} sqrt{3} right) approx 3.640919532942123602084014417112851989279ldots.$$
answered Jan 6 at 5:09
heropupheropup
63k66199
answered Jan 6 at 5:09
heropupheropup
63k66199
answered Jan 6 at 5:09
heropupheropup
63k66199
answered Jan 6 at 5:09
heropupheropup
63k66199
63k66199
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
add a comment |
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
1
1
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
$begingroup$
It is quite tedious (not simple, as you wrote) but we can get the antiderivatives and then the integrals appearing befor the result. This is a nice solution ! $to +1$. Cheers.
$endgroup$
– Claude Leibovici
Jan 6 at 7:00
add a comment |
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$begingroup$
In your link, integrand was just $1-x^2-y^2,$ not as in the integral of the question.
$endgroup$
– coffeemath
Jan 6 at 4:16
1
$begingroup$
@coffeemath It is just a link to the widget.
$endgroup$
– TED
Jan 6 at 4:19
$begingroup$
TED-- Did it give an "exact" value (closed form) or only an approximation $3.64092$? I ask since it may be there is no closed form-- WA usually finds one if it exists, I'm not familiar with Widget, which may only be a small subset of WA.
$endgroup$
– coffeemath
Jan 6 at 4:26
2
$begingroup$
@coffeemath It doesn't give closed form. I used the widget just to sort of know what value to expect from the integral.
$endgroup$
– TED
Jan 6 at 4:54
1
$begingroup$
The value does have an elementary closed form, but it is tedious. I am in the process of providing a proof.
$endgroup$
– heropup
Jan 6 at 5:01