Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$












0












$begingroup$


Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$



My work



Changing to cylindrical coordinates



For Paraboloid



$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$



For the cylinder



$$x^2+y^2=2ay;;;implies r=2a sin theta$$



Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$



Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$






multiple-integral







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$endgroup$












  • $begingroup$
    Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
    $endgroup$
    – Graham Kemp
    Aug 19 '17 at 9:25
















0












$begingroup$


Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$



My work



Changing to cylindrical coordinates



For Paraboloid



$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$



For the cylinder



$$x^2+y^2=2ay;;;implies r=2a sin theta$$



Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$



Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$






multiple-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
    $endgroup$
    – Graham Kemp
    Aug 19 '17 at 9:25














0












0








0


0



$begingroup$


Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$



My work



Changing to cylindrical coordinates



For Paraboloid



$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$



For the cylinder



$$x^2+y^2=2ay;;;implies r=2a sin theta$$



Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$



Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$






multiple-integral







share|cite|improve this question











$endgroup$




Find the volume bounded by the paraboloid $x^2+y^2=az$, the cylinder $x^2+y^2=2ay$ and the plane $z=0$



My work



Changing to cylindrical coordinates



For Paraboloid



$$x^2+y^2=az;;;implies r^2=az;;;implies z=frac{r^2}{a}$$



For the cylinder



$$x^2+y^2=2ay;;;implies r=2a sin theta$$



Since this volume lies only in first two quadrants $theta$ goes from $0$ to $pi$



Volume=$int_{theta=0}^{pi}int_{r=0}^{2asin theta}int_{z=0}^{r^2/a}r dr dtheta dz$






multiple-integral




multiple-integral






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share|cite|improve this question













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share|cite|improve this question








edited Aug 22 '17 at 0:37







user467745

















asked Aug 19 '17 at 9:00









user467745user467745

239112




239112












  • $begingroup$
    Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
    $endgroup$
    – Graham Kemp
    Aug 19 '17 at 9:25


















  • $begingroup$
    Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
    $endgroup$
    – Graham Kemp
    Aug 19 '17 at 9:25
















$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25




$begingroup$
Well, $yin[0..2a], xin[-sqrt{2ay-y^2}..sqrt{2ay-y^2}], zin[0..(x^2+y^2)/a]$ but a change of variables might be better.
$endgroup$
– Graham Kemp
Aug 19 '17 at 9:25










1 Answer
1






active

oldest

votes


















0












$begingroup$

Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, it does look much easier. But what would be the limits in cartesian coords?
    $endgroup$
    – user467745
    Aug 19 '17 at 9:21










  • $begingroup$
    @user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
    $endgroup$
    – Surb
    Aug 19 '17 at 9:25










  • $begingroup$
    Just to cross check-the answer would be $pi a^3$, right?
    $endgroup$
    – user467745
    Aug 19 '17 at 10:27












  • $begingroup$
    @user467745: I'm not a computer ;-)
    $endgroup$
    – Surb
    Aug 19 '17 at 10:35










  • $begingroup$
    Did I imply that? I am sorry. Thanks for your time
    $endgroup$
    – user467745
    Aug 19 '17 at 10:53












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, it does look much easier. But what would be the limits in cartesian coords?
    $endgroup$
    – user467745
    Aug 19 '17 at 9:21










  • $begingroup$
    @user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
    $endgroup$
    – Surb
    Aug 19 '17 at 9:25










  • $begingroup$
    Just to cross check-the answer would be $pi a^3$, right?
    $endgroup$
    – user467745
    Aug 19 '17 at 10:27












  • $begingroup$
    @user467745: I'm not a computer ;-)
    $endgroup$
    – Surb
    Aug 19 '17 at 10:35










  • $begingroup$
    Did I imply that? I am sorry. Thanks for your time
    $endgroup$
    – user467745
    Aug 19 '17 at 10:53
















0












$begingroup$

Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, it does look much easier. But what would be the limits in cartesian coords?
    $endgroup$
    – user467745
    Aug 19 '17 at 9:21










  • $begingroup$
    @user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
    $endgroup$
    – Surb
    Aug 19 '17 at 9:25










  • $begingroup$
    Just to cross check-the answer would be $pi a^3$, right?
    $endgroup$
    – user467745
    Aug 19 '17 at 10:27












  • $begingroup$
    @user467745: I'm not a computer ;-)
    $endgroup$
    – Surb
    Aug 19 '17 at 10:35










  • $begingroup$
    Did I imply that? I am sorry. Thanks for your time
    $endgroup$
    – user467745
    Aug 19 '17 at 10:53














0












0








0





$begingroup$

Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$






share|cite|improve this answer











$endgroup$



Using cylinder coordinate is the best way to solve this problem :
$$left{(rcostheta ,rsintheta +a ,z)mid theta in [0,2pi], rin[0,a], zin left[0,frac{2r^2+2arsin theta+a^2}{a}right]right}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 19 '17 at 15:43

























answered Aug 19 '17 at 9:15









SurbSurb

38.4k94478




38.4k94478












  • $begingroup$
    Yes, it does look much easier. But what would be the limits in cartesian coords?
    $endgroup$
    – user467745
    Aug 19 '17 at 9:21










  • $begingroup$
    @user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
    $endgroup$
    – Surb
    Aug 19 '17 at 9:25










  • $begingroup$
    Just to cross check-the answer would be $pi a^3$, right?
    $endgroup$
    – user467745
    Aug 19 '17 at 10:27












  • $begingroup$
    @user467745: I'm not a computer ;-)
    $endgroup$
    – Surb
    Aug 19 '17 at 10:35










  • $begingroup$
    Did I imply that? I am sorry. Thanks for your time
    $endgroup$
    – user467745
    Aug 19 '17 at 10:53


















  • $begingroup$
    Yes, it does look much easier. But what would be the limits in cartesian coords?
    $endgroup$
    – user467745
    Aug 19 '17 at 9:21










  • $begingroup$
    @user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
    $endgroup$
    – Surb
    Aug 19 '17 at 9:25










  • $begingroup$
    Just to cross check-the answer would be $pi a^3$, right?
    $endgroup$
    – user467745
    Aug 19 '17 at 10:27












  • $begingroup$
    @user467745: I'm not a computer ;-)
    $endgroup$
    – Surb
    Aug 19 '17 at 10:35










  • $begingroup$
    Did I imply that? I am sorry. Thanks for your time
    $endgroup$
    – user467745
    Aug 19 '17 at 10:53
















$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21




$begingroup$
Yes, it does look much easier. But what would be the limits in cartesian coords?
$endgroup$
– user467745
Aug 19 '17 at 9:21












$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25




$begingroup$
@user467745 : Something as $yin [0,2a]$, $xin [-sqrt{2a-(y-1)^2},sqrt{2a-(y-1)^2}]$ and $zin [0, frac{x^2}{a}+frac{y^2}{a}]$.
$endgroup$
– Surb
Aug 19 '17 at 9:25












$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27






$begingroup$
Just to cross check-the answer would be $pi a^3$, right?
$endgroup$
– user467745
Aug 19 '17 at 10:27














$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35




$begingroup$
@user467745: I'm not a computer ;-)
$endgroup$
– Surb
Aug 19 '17 at 10:35












$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53




$begingroup$
Did I imply that? I am sorry. Thanks for your time
$endgroup$
– user467745
Aug 19 '17 at 10:53


















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