## triple integral/cylindrical coordinates: 1+x/(x^2+y^2)^(1/2)

Limits, differentiation, related rates, integration, trig integrals, etc.
purell
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Joined: Thu Apr 16, 2009 1:22 am
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### triple integral/cylindrical coordinates: 1+x/(x^2+y^2)^(1/2)

i have to find the triple integral of 1+x/(x^2+y^2)^(1/2) dx dy dz where S is the solid bounded by the paraboloids z=x^2+y^2 and z=1-x^2-y^2.

this is as far as i have gotten:
F(r,theta,z)=(r*cos(theta),r*sin(theta),z)
abs(det(JF))=r

so now i have the triple integral of r(1+cos(theta)) dz dr dtheta.

but how do i find the limits of integration?

Martingale
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Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
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### Re: triple integral/cylindrical coordinates

i have to find the triple integral of 1+x/(x^2+y^2)^(1/2) dx dy dz where S is the solid bounded by the paraboloids z=x^2+y^2 and z=1-x^2-y^2.

this is as far as i have gotten:
F(r,theta,z)=(r*cos(theta),r*sin(theta),z)
abs(det(JF))=r

so now i have the triple integral of r(1+cos(theta)) dz dr dtheta.

but how do i find the limits of integration?
if your integrand is $1+\frac{x}{\sqrt{x^2+y^2}$

then I like your conversion to cylindrical coordinates...

we have $z=x^2+y^2$ and $z=1-x^2-y^2$

so

$z=r^2$ and $z=1-r^2$

these are the upper and lower bounds for $z$ so we have

$r^2\leq z\leq1-r^2$

where the two paraboloids meet $r^2=1-r^2$ so $r=\frac{1}{\sqrt{2}}$

is the maximum $r$ can be.

so we should be able to write the integral as

$\int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r^2}^{1-r^2}r(1+\cos(\theta))dzdrd\theta$

purell
Posts: 2
Joined: Thu Apr 16, 2009 1:22 am
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thanks so much!