$int_{x^2+y^2+z^2 leq 1}frac{dx,dy,dz}{x^2+y^2+(z-2)^2}$












1












$begingroup$


I'm trying to calculate the integral $$int_{x^2+y^2+z^2 leq 1}frac{dx,dy,dz}{x^2+y^2+(z-2)^2}.$$



I've tried in two methods:



Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.



The other method was to try shifted spherical coordinates, $x = rsin(theta)cos(phi), y=rsin(theta)sin(phi)$ but $z-2 = rcos(theta)$.



Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < theta < pi$ and $0 < phi < 2pi$, I think, but the limits on $r$ are harder.

At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4rcos(theta)+4 = 1$, so we need to have $r^2+4rcos(theta) + 3 = 0$.



This happens when $$r = frac{-4cos(theta)pmsqrt{16cos^2(theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.



The unfun way:



$int_{x^2+y^2+z^2 leq 1}frac{dxdydz}{x^2+y^2+(z-2)^2} = int_{0}^{1}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dphi dtheta dr = 2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr $



Use $u-$substitution $u=cos(theta)$ to get:



$2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr =-2pi int_{0}^{1}int_{1}^{-1}frac{r^2}{r^2-4ru+4}dudr = 2piint_{0}^{1}int_{-1}^{1}frac{r^2}{r^2-4ru+r}dudr = 2piint_{0}^{1}r^2[frac{ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -frac{pi}{2}int_{0}^{1}r(ln(r^2-4r+4) - ln(r^2+4r+4))dr=-piint_{0}^{1}r(ln(|r-2|)-ln(r+2))dr$



Is this the right way? seems very unpleasant...






integration change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
    $endgroup$
    – Masacroso
    Jan 2 at 20:41










  • $begingroup$
    A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
    $endgroup$
    – jmerry
    Jan 2 at 21:19
















1












$begingroup$


I'm trying to calculate the integral $$int_{x^2+y^2+z^2 leq 1}frac{dx,dy,dz}{x^2+y^2+(z-2)^2}.$$



I've tried in two methods:



Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.



The other method was to try shifted spherical coordinates, $x = rsin(theta)cos(phi), y=rsin(theta)sin(phi)$ but $z-2 = rcos(theta)$.



Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < theta < pi$ and $0 < phi < 2pi$, I think, but the limits on $r$ are harder.

At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4rcos(theta)+4 = 1$, so we need to have $r^2+4rcos(theta) + 3 = 0$.



This happens when $$r = frac{-4cos(theta)pmsqrt{16cos^2(theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.



The unfun way:



$int_{x^2+y^2+z^2 leq 1}frac{dxdydz}{x^2+y^2+(z-2)^2} = int_{0}^{1}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dphi dtheta dr = 2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr $



Use $u-$substitution $u=cos(theta)$ to get:



$2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr =-2pi int_{0}^{1}int_{1}^{-1}frac{r^2}{r^2-4ru+4}dudr = 2piint_{0}^{1}int_{-1}^{1}frac{r^2}{r^2-4ru+r}dudr = 2piint_{0}^{1}r^2[frac{ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -frac{pi}{2}int_{0}^{1}r(ln(r^2-4r+4) - ln(r^2+4r+4))dr=-piint_{0}^{1}r(ln(|r-2|)-ln(r+2))dr$



Is this the right way? seems very unpleasant...






integration change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
    $endgroup$
    – Masacroso
    Jan 2 at 20:41










  • $begingroup$
    A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
    $endgroup$
    – jmerry
    Jan 2 at 21:19














1












1








1





$begingroup$


I'm trying to calculate the integral $$int_{x^2+y^2+z^2 leq 1}frac{dx,dy,dz}{x^2+y^2+(z-2)^2}.$$



I've tried in two methods:



Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.



The other method was to try shifted spherical coordinates, $x = rsin(theta)cos(phi), y=rsin(theta)sin(phi)$ but $z-2 = rcos(theta)$.



Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < theta < pi$ and $0 < phi < 2pi$, I think, but the limits on $r$ are harder.

At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4rcos(theta)+4 = 1$, so we need to have $r^2+4rcos(theta) + 3 = 0$.



This happens when $$r = frac{-4cos(theta)pmsqrt{16cos^2(theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.



The unfun way:



$int_{x^2+y^2+z^2 leq 1}frac{dxdydz}{x^2+y^2+(z-2)^2} = int_{0}^{1}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dphi dtheta dr = 2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr $



Use $u-$substitution $u=cos(theta)$ to get:



$2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr =-2pi int_{0}^{1}int_{1}^{-1}frac{r^2}{r^2-4ru+4}dudr = 2piint_{0}^{1}int_{-1}^{1}frac{r^2}{r^2-4ru+r}dudr = 2piint_{0}^{1}r^2[frac{ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -frac{pi}{2}int_{0}^{1}r(ln(r^2-4r+4) - ln(r^2+4r+4))dr=-piint_{0}^{1}r(ln(|r-2|)-ln(r+2))dr$



Is this the right way? seems very unpleasant...






integration change-of-variable







share|cite|improve this question











$endgroup$




I'm trying to calculate the integral $$int_{x^2+y^2+z^2 leq 1}frac{dx,dy,dz}{x^2+y^2+(z-2)^2}.$$



I've tried in two methods:



Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.



The other method was to try shifted spherical coordinates, $x = rsin(theta)cos(phi), y=rsin(theta)sin(phi)$ but $z-2 = rcos(theta)$.



Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < theta < pi$ and $0 < phi < 2pi$, I think, but the limits on $r$ are harder.

At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4rcos(theta)+4 = 1$, so we need to have $r^2+4rcos(theta) + 3 = 0$.



This happens when $$r = frac{-4cos(theta)pmsqrt{16cos^2(theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.



The unfun way:



$int_{x^2+y^2+z^2 leq 1}frac{dxdydz}{x^2+y^2+(z-2)^2} = int_{0}^{1}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dphi dtheta dr = 2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr $



Use $u-$substitution $u=cos(theta)$ to get:



$2pi int_{0}^{1}int_{0}^{pi}frac{r^2sin(theta)}{r^2-4rcos(theta)+4}dtheta dr =-2pi int_{0}^{1}int_{1}^{-1}frac{r^2}{r^2-4ru+4}dudr = 2piint_{0}^{1}int_{-1}^{1}frac{r^2}{r^2-4ru+r}dudr = 2piint_{0}^{1}r^2[frac{ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -frac{pi}{2}int_{0}^{1}r(ln(r^2-4r+4) - ln(r^2+4r+4))dr=-piint_{0}^{1}r(ln(|r-2|)-ln(r+2))dr$



Is this the right way? seems very unpleasant...






integration change-of-variable




integration change-of-variable






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 20:40







Oria Gruber

















asked Jan 2 at 20:27









Oria GruberOria Gruber

6,53232461




6,53232461












  • $begingroup$
    decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
    $endgroup$
    – Masacroso
    Jan 2 at 20:41










  • $begingroup$
    A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
    $endgroup$
    – jmerry
    Jan 2 at 21:19


















  • $begingroup$
    decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
    $endgroup$
    – Masacroso
    Jan 2 at 20:41










  • $begingroup$
    A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
    $endgroup$
    – jmerry
    Jan 2 at 21:19
















$begingroup$
decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
$endgroup$
– Masacroso
Jan 2 at 20:41




$begingroup$
decompose the open ball (the region of integration) in circular sectors. See the answer of @Sameer
$endgroup$
– Masacroso
Jan 2 at 20:41












$begingroup$
A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
$endgroup$
– jmerry
Jan 2 at 21:19




$begingroup$
A technical issue - that all-TeX title disables the "open in a new window/tab" right-click menu. Please edit it to include one or more words.
$endgroup$
– jmerry
Jan 2 at 21:19










1 Answer
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active

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1












$begingroup$

Consider the integral $$I(a, b) = int_{x^2 + y^2 le a} frac{dx , dy, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = int_{0}^{2pi} int_{0}^{sqrt{a}} frac{r , dr, dtheta}{r^2 + b} = pi ln(a/b + 1)$$ Your desired integral is $$int_{-1}^{1} I(1-z^2, (z-2)^2) , dz = pi int_{-1}^{1} lnleft(frac{1-z^2 + (z-2)^2}{(z-2)^{2}} right), dz = pi int_{-1}^{1} ln(5-4z) - 2ln(|z-2|) , dz$$ which one can compute by elementary techniques (e.g. integration by parts).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's Cavalieri's principle right? Should have thought of that trick.
    $endgroup$
    – Oria Gruber
    Jan 2 at 20:42










  • $begingroup$
    What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
    $endgroup$
    – Oria Gruber
    Jan 2 at 21:12











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1 Answer
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1












$begingroup$

Consider the integral $$I(a, b) = int_{x^2 + y^2 le a} frac{dx , dy, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = int_{0}^{2pi} int_{0}^{sqrt{a}} frac{r , dr, dtheta}{r^2 + b} = pi ln(a/b + 1)$$ Your desired integral is $$int_{-1}^{1} I(1-z^2, (z-2)^2) , dz = pi int_{-1}^{1} lnleft(frac{1-z^2 + (z-2)^2}{(z-2)^{2}} right), dz = pi int_{-1}^{1} ln(5-4z) - 2ln(|z-2|) , dz$$ which one can compute by elementary techniques (e.g. integration by parts).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's Cavalieri's principle right? Should have thought of that trick.
    $endgroup$
    – Oria Gruber
    Jan 2 at 20:42










  • $begingroup$
    What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
    $endgroup$
    – Oria Gruber
    Jan 2 at 21:12
















1












$begingroup$

Consider the integral $$I(a, b) = int_{x^2 + y^2 le a} frac{dx , dy, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = int_{0}^{2pi} int_{0}^{sqrt{a}} frac{r , dr, dtheta}{r^2 + b} = pi ln(a/b + 1)$$ Your desired integral is $$int_{-1}^{1} I(1-z^2, (z-2)^2) , dz = pi int_{-1}^{1} lnleft(frac{1-z^2 + (z-2)^2}{(z-2)^{2}} right), dz = pi int_{-1}^{1} ln(5-4z) - 2ln(|z-2|) , dz$$ which one can compute by elementary techniques (e.g. integration by parts).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That's Cavalieri's principle right? Should have thought of that trick.
    $endgroup$
    – Oria Gruber
    Jan 2 at 20:42










  • $begingroup$
    What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
    $endgroup$
    – Oria Gruber
    Jan 2 at 21:12














1












1








1





$begingroup$

Consider the integral $$I(a, b) = int_{x^2 + y^2 le a} frac{dx , dy, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = int_{0}^{2pi} int_{0}^{sqrt{a}} frac{r , dr, dtheta}{r^2 + b} = pi ln(a/b + 1)$$ Your desired integral is $$int_{-1}^{1} I(1-z^2, (z-2)^2) , dz = pi int_{-1}^{1} lnleft(frac{1-z^2 + (z-2)^2}{(z-2)^{2}} right), dz = pi int_{-1}^{1} ln(5-4z) - 2ln(|z-2|) , dz$$ which one can compute by elementary techniques (e.g. integration by parts).






share|cite|improve this answer









$endgroup$



Consider the integral $$I(a, b) = int_{x^2 + y^2 le a} frac{dx , dy, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = int_{0}^{2pi} int_{0}^{sqrt{a}} frac{r , dr, dtheta}{r^2 + b} = pi ln(a/b + 1)$$ Your desired integral is $$int_{-1}^{1} I(1-z^2, (z-2)^2) , dz = pi int_{-1}^{1} lnleft(frac{1-z^2 + (z-2)^2}{(z-2)^{2}} right), dz = pi int_{-1}^{1} ln(5-4z) - 2ln(|z-2|) , dz$$ which one can compute by elementary techniques (e.g. integration by parts).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 20:38









Sameer KailasaSameer Kailasa

5,57321843




5,57321843












  • $begingroup$
    That's Cavalieri's principle right? Should have thought of that trick.
    $endgroup$
    – Oria Gruber
    Jan 2 at 20:42










  • $begingroup$
    What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
    $endgroup$
    – Oria Gruber
    Jan 2 at 21:12


















  • $begingroup$
    That's Cavalieri's principle right? Should have thought of that trick.
    $endgroup$
    – Oria Gruber
    Jan 2 at 20:42










  • $begingroup$
    What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
    $endgroup$
    – Oria Gruber
    Jan 2 at 21:12
















$begingroup$
That's Cavalieri's principle right? Should have thought of that trick.
$endgroup$
– Oria Gruber
Jan 2 at 20:42




$begingroup$
That's Cavalieri's principle right? Should have thought of that trick.
$endgroup$
– Oria Gruber
Jan 2 at 20:42












$begingroup$
What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
$endgroup$
– Oria Gruber
Jan 2 at 21:12




$begingroup$
What would you do if the integrand was $(z-frac{1}{2})^2$ instead? You can't use Cavalieri now, it's now an improper integral.
$endgroup$
– Oria Gruber
Jan 2 at 21:12


















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