Mixing
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.
calculus integration multivariable-calculus definite-integrals area
asked 2 hours ago
NetUser5y62
389114
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.
calculus integration multivariable-calculus definite-integrals area
asked 2 hours ago
NetUser5y62
389114
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
1 hour ago
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.
calculus integration multivariable-calculus definite-integrals area
asked 2 hours ago
NetUser5y62
389114
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.
calculus integration multivariable-calculus definite-integrals area
calculus integration multivariable-calculus definite-integrals area
asked 2 hours ago
NetUser5y62
389114
asked 2 hours ago
NetUser5y62
389114
edited 50 mins ago
asked 2 hours ago
NetUser5y62
389114
asked 2 hours ago
NetUser5y62
389114
asked 2 hours ago
NetUser5y62
389114
389114
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago