Substitution for a double integral
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Show that
$$
4int_0^1 int_0^1 sqrt{1+x^2+y^2} , dy , dx = frac{8}{3} int_0^{pi /4} (1+ sec^2 theta )^{1.5} , dtheta - frac{2pi}{3}
$$
I have tried letting $1+x^2 = lambda^2$ , $y=lambda tan theta$ , $frac{dy}{dtheta} = lambda sec^2 theta$. After which
$$begin{aligned}
4int_0^1 int_{arctan (0/ lambda)}^{arctan (1/ lambda)} lambda^2 sec^3 theta , dtheta , dx &= 4int_0^1 lambda^2 left[ 0.5 sec theta tan theta + 0.5 ln (sec theta + tan theta) right]_{arctan (0/ lambda)}^{arctan (1/ lambda)}, dx
end{aligned} $$
And then I was convinced this was not the way to go.
integration multivariable-calculus definite-integrals
asked Nov 19 at 8:17
NetUser5y62
393114
add a comment |
up vote
0
down vote
favorite
Show that
$$
4int_0^1 int_0^1 sqrt{1+x^2+y^2} , dy , dx = frac{8}{3} int_0^{pi /4} (1+ sec^2 theta )^{1.5} , dtheta - frac{2pi}{3}
$$
I have tried letting $1+x^2 = lambda^2$ , $y=lambda tan theta$ , $frac{dy}{dtheta} = lambda sec^2 theta$. After which
$$begin{aligned}
4int_0^1 int_{arctan (0/ lambda)}^{arctan (1/ lambda)} lambda^2 sec^3 theta , dtheta , dx &= 4int_0^1 lambda^2 left[ 0.5 sec theta tan theta + 0.5 ln (sec theta + tan theta) right]_{arctan (0/ lambda)}^{arctan (1/ lambda)}, dx
end{aligned} $$
And then I was convinced this was not the way to go.
integration multivariable-calculus definite-integrals
asked Nov 19 at 8:17
NetUser5y62
393114
I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that
$$
4int_0^1 int_0^1 sqrt{1+x^2+y^2} , dy , dx = frac{8}{3} int_0^{pi /4} (1+ sec^2 theta )^{1.5} , dtheta - frac{2pi}{3}
$$
I have tried letting $1+x^2 = lambda^2$ , $y=lambda tan theta$ , $frac{dy}{dtheta} = lambda sec^2 theta$. After which
$$begin{aligned}
4int_0^1 int_{arctan (0/ lambda)}^{arctan (1/ lambda)} lambda^2 sec^3 theta , dtheta , dx &= 4int_0^1 lambda^2 left[ 0.5 sec theta tan theta + 0.5 ln (sec theta + tan theta) right]_{arctan (0/ lambda)}^{arctan (1/ lambda)}, dx
end{aligned} $$
And then I was convinced this was not the way to go.
integration multivariable-calculus definite-integrals
asked Nov 19 at 8:17
NetUser5y62
393114
Show that
$$
4int_0^1 int_0^1 sqrt{1+x^2+y^2} , dy , dx = frac{8}{3} int_0^{pi /4} (1+ sec^2 theta )^{1.5} , dtheta - frac{2pi}{3}
$$
I have tried letting $1+x^2 = lambda^2$ , $y=lambda tan theta$ , $frac{dy}{dtheta} = lambda sec^2 theta$. After which
$$begin{aligned}
4int_0^1 int_{arctan (0/ lambda)}^{arctan (1/ lambda)} lambda^2 sec^3 theta , dtheta , dx &= 4int_0^1 lambda^2 left[ 0.5 sec theta tan theta + 0.5 ln (sec theta + tan theta) right]_{arctan (0/ lambda)}^{arctan (1/ lambda)}, dx
end{aligned} $$
And then I was convinced this was not the way to go.
integration multivariable-calculus definite-integrals
integration multivariable-calculus definite-integrals
asked Nov 19 at 8:17
NetUser5y62
393114
asked Nov 19 at 8:17
NetUser5y62
393114
edited Nov 19 at 11:04
asked Nov 19 at 8:17
NetUser5y62
393114
asked Nov 19 at 8:17
NetUser5y62
393114
asked Nov 19 at 8:17
NetUser5y62
393114
393114
I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38
add a comment |
I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38
I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8int_0^1 int_0^x f(x,y) ; dy ; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $rcos theta =1$ or $r=sec theta.$ So in polar, the integral is
$$8int_0^{pi/4} int_0^{sec theta} sqrt{1+r^2}; r ; dr ; dtheta.$$
Then things are straightforward.
answered Nov 19 at 11:53
B. Goddard
18.2k21340
add a comment |
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 function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8int_0^1 int_0^x f(x,y) ; dy ; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $rcos theta =1$ or $r=sec theta.$ So in polar, the integral is
$$8int_0^{pi/4} int_0^{sec theta} sqrt{1+r^2}; r ; dr ; dtheta.$$
Then things are straightforward.
answered Nov 19 at 11:53
B. Goddard
18.2k21340
add a comment |
up vote
2
down vote
accepted
The function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8int_0^1 int_0^x f(x,y) ; dy ; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $rcos theta =1$ or $r=sec theta.$ So in polar, the integral is
$$8int_0^{pi/4} int_0^{sec theta} sqrt{1+r^2}; r ; dr ; dtheta.$$
Then things are straightforward.
answered Nov 19 at 11:53
B. Goddard
18.2k21340
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8int_0^1 int_0^x f(x,y) ; dy ; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $rcos theta =1$ or $r=sec theta.$ So in polar, the integral is
$$8int_0^{pi/4} int_0^{sec theta} sqrt{1+r^2}; r ; dr ; dtheta.$$
Then things are straightforward.
answered Nov 19 at 11:53
B. Goddard
18.2k21340
The function and the region are symmetric about the line $y=x$, so you can re-write your integral as
$$8int_0^1 int_0^x f(x,y) ; dy ; dx.$$
So you're integrating over the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$
The vertical line $x=1$ has polar equation $rcos theta =1$ or $r=sec theta.$ So in polar, the integral is
$$8int_0^{pi/4} int_0^{sec theta} sqrt{1+r^2}; r ; dr ; dtheta.$$
Then things are straightforward.
answered Nov 19 at 11:53
B. Goddard
18.2k21340
answered Nov 19 at 11:53
B. Goddard
18.2k21340
answered Nov 19 at 11:53
B. Goddard
18.2k21340
answered Nov 19 at 11:53
B. Goddard
18.2k21340
18.2k21340
add a comment |
add a comment |
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I think you should switch to polar coordinates and then integrate w.r.t. $r=sqrt {1+x^{2}}$ first. Consider integral over the four quadrants separately.
– Kavi Rama Murthy
Nov 19 at 8:23
Describing $[0, 1]^2$ in polar coordinates doesn't seems like fun though - have you tried computing the integral over $y$ explicitly?
– Stockfish
Nov 19 at 11:26
I am clueless. I did my part for the research and found this: mathproofs.blogspot.com/2005/07/mapping-square-to-circle.html , but I am still clueless.
– NetUser5y62
Nov 19 at 11:38