integrate $int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$

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Question: Integrate the function $f(x,y,z)=y$ over the region R bounded by the plane $x+y+z=2$, the cylinder $x^2+z^2=1$, and the plane $y=0$.



My solution: since, $0<=y<=2-x-z$ and $x^2+z^2=1$



so let $x=rcostheta$ and $z=r sintheta$ and then I get the integration:



$$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$
I feel like I'm making this very complicated. Am i doing something wrong?




integration multivariable-calculus




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  • You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
    – Ted Shifrin
    Jul 21 at 0:38














up vote
0
down vote

favorite












Question: Integrate the function $f(x,y,z)=y$ over the region R bounded by the plane $x+y+z=2$, the cylinder $x^2+z^2=1$, and the plane $y=0$.



My solution: since, $0<=y<=2-x-z$ and $x^2+z^2=1$



so let $x=rcostheta$ and $z=r sintheta$ and then I get the integration:



$$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$
I feel like I'm making this very complicated. Am i doing something wrong?




integration multivariable-calculus




share|cite|improve this question



















  • You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
    – Ted Shifrin
    Jul 21 at 0:38












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Question: Integrate the function $f(x,y,z)=y$ over the region R bounded by the plane $x+y+z=2$, the cylinder $x^2+z^2=1$, and the plane $y=0$.



My solution: since, $0<=y<=2-x-z$ and $x^2+z^2=1$



so let $x=rcostheta$ and $z=r sintheta$ and then I get the integration:



$$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$
I feel like I'm making this very complicated. Am i doing something wrong?




integration multivariable-calculus




share|cite|improve this question











Question: Integrate the function $f(x,y,z)=y$ over the region R bounded by the plane $x+y+z=2$, the cylinder $x^2+z^2=1$, and the plane $y=0$.



My solution: since, $0<=y<=2-x-z$ and $x^2+z^2=1$



so let $x=rcostheta$ and $z=r sintheta$ and then I get the integration:



$$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$
I feel like I'm making this very complicated. Am i doing something wrong?





integration multivariable-calculus





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 21 at 0:08









thepanda

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  • You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
    – Ted Shifrin
    Jul 21 at 0:38
















  • You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
    – Ted Shifrin
    Jul 21 at 0:38















You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
– Ted Shifrin
Jul 21 at 0:38




You should say $x^2+z^2le 1$, but otherwise everything is fine. Great idea to use cylindrical coordinates based on the $xz$-plane.
– Ted Shifrin
Jul 21 at 0:38










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Your integral $$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$ should have been $$int_0^2 piint_0^1 int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$



Otherwise it is OK.






share|cite|improve this answer





















  • thanks! I didn't even notice the mistake
    – thepanda
    Jul 21 at 0:45










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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








up vote
1
down vote



accepted










Your integral $$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$ should have been $$int_0^2 piint_0^1 int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$



Otherwise it is OK.






share|cite|improve this answer





















  • thanks! I didn't even notice the mistake
    – thepanda
    Jul 21 at 0:45














up vote
1
down vote



accepted










Your integral $$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$ should have been $$int_0^2 piint_0^1 int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$



Otherwise it is OK.






share|cite|improve this answer





















  • thanks! I didn't even notice the mistake
    – thepanda
    Jul 21 at 0:45












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Your integral $$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$ should have been $$int_0^2 piint_0^1 int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$



Otherwise it is OK.






share|cite|improve this answer













Your integral $$int_0^1 int_0^2 pi int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$ should have been $$int_0^2 piint_0^1 int_0^2-rcostheta - rsintheta yr,dy,dr,dtheta$$



Otherwise it is OK.







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share|cite|improve this answer











answered Jul 21 at 0:43









Mohammad Riazi-Kermani

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  • thanks! I didn't even notice the mistake
    – thepanda
    Jul 21 at 0:45
















  • thanks! I didn't even notice the mistake
    – thepanda
    Jul 21 at 0:45















thanks! I didn't even notice the mistake
– thepanda
Jul 21 at 0:45




thanks! I didn't even notice the mistake
– thepanda
Jul 21 at 0:45












 

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