how to find the volume of intersection between eleptic cylinder and cylinder?












0












$begingroup$



let C be the eleptic cylinder C = {$frac{y^2}{4} + z^2leq 1$}.



let D be the cylinder {$x^2 + z^2 leq 1$}
find the volume of intersection of C and D




this is what i did :
$$V = int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}int_{x=-sqrt{1-z^2}}^{x=sqrt{1-z^2}} 1{ dx}{ dy}{ dz} $$



$$= int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}2sqrt{1-z^2}{ dy}{ dz}=$$
$$= int_{z=-1}^{z=1}2sqrt{1-z^2}*2sqrt{4-4z^2} { dz}= int_{z=-1}^{z=1}4sqrt{4z^4-8z^2+4} { dz}=int_{z=-1}^{z=1}4sqrt{(2z^2 -2)^2} { dz} = 10frac{2}{3}$$



why did i get negative volume ?
did i do it correctly ? or am i missing something






integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    correct edited *
    $endgroup$
    – Mather
    Dec 15 '18 at 15:12












  • $begingroup$
    The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:15










  • $begingroup$
    i have posted extended calculation
    $endgroup$
    – Mather
    Dec 15 '18 at 15:28










  • $begingroup$
    I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:31












  • $begingroup$
    the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
    $endgroup$
    – Mather
    Dec 15 '18 at 15:33


















0












$begingroup$



let C be the eleptic cylinder C = {$frac{y^2}{4} + z^2leq 1$}.



let D be the cylinder {$x^2 + z^2 leq 1$}
find the volume of intersection of C and D




this is what i did :
$$V = int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}int_{x=-sqrt{1-z^2}}^{x=sqrt{1-z^2}} 1{ dx}{ dy}{ dz} $$



$$= int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}2sqrt{1-z^2}{ dy}{ dz}=$$
$$= int_{z=-1}^{z=1}2sqrt{1-z^2}*2sqrt{4-4z^2} { dz}= int_{z=-1}^{z=1}4sqrt{4z^4-8z^2+4} { dz}=int_{z=-1}^{z=1}4sqrt{(2z^2 -2)^2} { dz} = 10frac{2}{3}$$



why did i get negative volume ?
did i do it correctly ? or am i missing something






integration multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    correct edited *
    $endgroup$
    – Mather
    Dec 15 '18 at 15:12












  • $begingroup$
    The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:15










  • $begingroup$
    i have posted extended calculation
    $endgroup$
    – Mather
    Dec 15 '18 at 15:28










  • $begingroup$
    I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:31












  • $begingroup$
    the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
    $endgroup$
    – Mather
    Dec 15 '18 at 15:33
















0












0








0





$begingroup$



let C be the eleptic cylinder C = {$frac{y^2}{4} + z^2leq 1$}.



let D be the cylinder {$x^2 + z^2 leq 1$}
find the volume of intersection of C and D




this is what i did :
$$V = int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}int_{x=-sqrt{1-z^2}}^{x=sqrt{1-z^2}} 1{ dx}{ dy}{ dz} $$



$$= int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}2sqrt{1-z^2}{ dy}{ dz}=$$
$$= int_{z=-1}^{z=1}2sqrt{1-z^2}*2sqrt{4-4z^2} { dz}= int_{z=-1}^{z=1}4sqrt{4z^4-8z^2+4} { dz}=int_{z=-1}^{z=1}4sqrt{(2z^2 -2)^2} { dz} = 10frac{2}{3}$$



why did i get negative volume ?
did i do it correctly ? or am i missing something






integration multivariable-calculus







share|cite|improve this question











$endgroup$





let C be the eleptic cylinder C = {$frac{y^2}{4} + z^2leq 1$}.



let D be the cylinder {$x^2 + z^2 leq 1$}
find the volume of intersection of C and D




this is what i did :
$$V = int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}int_{x=-sqrt{1-z^2}}^{x=sqrt{1-z^2}} 1{ dx}{ dy}{ dz} $$



$$= int_{z=-1}^{z=1}int_{y=-sqrt{4-4z^2}}^{y=sqrt{4-4z^2}}2sqrt{1-z^2}{ dy}{ dz}=$$
$$= int_{z=-1}^{z=1}2sqrt{1-z^2}*2sqrt{4-4z^2} { dz}= int_{z=-1}^{z=1}4sqrt{4z^4-8z^2+4} { dz}=int_{z=-1}^{z=1}4sqrt{(2z^2 -2)^2} { dz} = 10frac{2}{3}$$



why did i get negative volume ?
did i do it correctly ? or am i missing something






integration multivariable-calculus




integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 15:43







Mather

















asked Dec 15 '18 at 15:07









Mather Mather

3317




3317












  • $begingroup$
    correct edited *
    $endgroup$
    – Mather
    Dec 15 '18 at 15:12












  • $begingroup$
    The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:15










  • $begingroup$
    i have posted extended calculation
    $endgroup$
    – Mather
    Dec 15 '18 at 15:28










  • $begingroup$
    I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:31












  • $begingroup$
    the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
    $endgroup$
    – Mather
    Dec 15 '18 at 15:33




















  • $begingroup$
    correct edited *
    $endgroup$
    – Mather
    Dec 15 '18 at 15:12












  • $begingroup$
    The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:15










  • $begingroup$
    i have posted extended calculation
    $endgroup$
    – Mather
    Dec 15 '18 at 15:28










  • $begingroup$
    I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:31












  • $begingroup$
    the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
    $endgroup$
    – Mather
    Dec 15 '18 at 15:33


















$begingroup$
correct edited *
$endgroup$
– Mather
Dec 15 '18 at 15:12






$begingroup$
correct edited *
$endgroup$
– Mather
Dec 15 '18 at 15:12














$begingroup$
The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
$endgroup$
– saulspatz
Dec 15 '18 at 15:15




$begingroup$
The integral looks right to me, and you certainly shouldn't get a negative answer. Please post your calculations, so we can pinpoint the error.
$endgroup$
– saulspatz
Dec 15 '18 at 15:15












$begingroup$
i have posted extended calculation
$endgroup$
– Mather
Dec 15 '18 at 15:28




$begingroup$
i have posted extended calculation
$endgroup$
– Mather
Dec 15 '18 at 15:28












$begingroup$
I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
$endgroup$
– saulspatz
Dec 15 '18 at 15:31






$begingroup$
I agree up to the very last equal sign. How did you do the last integral? I'm guessing you simplified $sqrt{z^2-1}$ as $z^2-1$ instead of $1-z^2$.
$endgroup$
– saulspatz
Dec 15 '18 at 15:31














$begingroup$
the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
$endgroup$
– Mather
Dec 15 '18 at 15:33






$begingroup$
the squrt of $(2z - 2)^2 $ is $ 2z-2 $ so doing the integral it is $ z^3/3 - 2z $
$endgroup$
– Mather
Dec 15 '18 at 15:33












1 Answer
1






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oldest

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1












$begingroup$

You are correct up to the last integral. Recall that $sqrt{z^2}=|z|,$ so $sqrt{(z^2-1)^2}=|z^2-1|$ When $-1le zle1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh i got it now thank you
    $endgroup$
    – Mather
    Dec 15 '18 at 15:38










  • $begingroup$
    It was my pleasure.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:40











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are correct up to the last integral. Recall that $sqrt{z^2}=|z|,$ so $sqrt{(z^2-1)^2}=|z^2-1|$ When $-1le zle1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh i got it now thank you
    $endgroup$
    – Mather
    Dec 15 '18 at 15:38










  • $begingroup$
    It was my pleasure.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:40
















1












$begingroup$

You are correct up to the last integral. Recall that $sqrt{z^2}=|z|,$ so $sqrt{(z^2-1)^2}=|z^2-1|$ When $-1le zle1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    oh i got it now thank you
    $endgroup$
    – Mather
    Dec 15 '18 at 15:38










  • $begingroup$
    It was my pleasure.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:40














1












1








1





$begingroup$

You are correct up to the last integral. Recall that $sqrt{z^2}=|z|,$ so $sqrt{(z^2-1)^2}=|z^2-1|$ When $-1le zle1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.






share|cite|improve this answer









$endgroup$



You are correct up to the last integral. Recall that $sqrt{z^2}=|z|,$ so $sqrt{(z^2-1)^2}=|z^2-1|$ When $-1le zle1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 15:37









saulspatzsaulspatz

15.1k31330




15.1k31330












  • $begingroup$
    oh i got it now thank you
    $endgroup$
    – Mather
    Dec 15 '18 at 15:38










  • $begingroup$
    It was my pleasure.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:40


















  • $begingroup$
    oh i got it now thank you
    $endgroup$
    – Mather
    Dec 15 '18 at 15:38










  • $begingroup$
    It was my pleasure.
    $endgroup$
    – saulspatz
    Dec 15 '18 at 15:40
















$begingroup$
oh i got it now thank you
$endgroup$
– Mather
Dec 15 '18 at 15:38




$begingroup$
oh i got it now thank you
$endgroup$
– Mather
Dec 15 '18 at 15:38












$begingroup$
It was my pleasure.
$endgroup$
– saulspatz
Dec 15 '18 at 15:40




$begingroup$
It was my pleasure.
$endgroup$
– saulspatz
Dec 15 '18 at 15:40


















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