Triple integral of portion of cone (cylindrical polar coordinates)?












3














$V$ is the portion of the cone $z=sqrt{x^2+y^2};$ for $; xgeq 0.$
Find $$iiintlimits_{V} xe^{-z} dV.$$
I am trying to solve this question. The answer is supposed to be just $4.$



I have worked out the limits as



$0leq z leq infty;$ and
$;-pi/2leq theta leq pi/2;$ and $;0leq R leq z.$ What am I doing wrong?






calculus multivariable-calculus







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  • 2




    Please check if I've edited it correctly.
    – user376343
    Dec 7 at 21:59












  • yes, thankyou. For future, how does one use latex here?
    – Pumpkinpeach
    Dec 7 at 22:12










  • We are curious to know what the issue was with the evaluation.
    – gimusi
    Dec 7 at 22:13










  • @Pumpkinpeach Refer to MathJax basic tutorial and quick reference
    – gimusi
    Dec 7 at 22:14










  • @gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
    – Pumpkinpeach
    Dec 7 at 22:17
















3














$V$ is the portion of the cone $z=sqrt{x^2+y^2};$ for $; xgeq 0.$
Find $$iiintlimits_{V} xe^{-z} dV.$$
I am trying to solve this question. The answer is supposed to be just $4.$



I have worked out the limits as



$0leq z leq infty;$ and
$;-pi/2leq theta leq pi/2;$ and $;0leq R leq z.$ What am I doing wrong?






calculus multivariable-calculus







share|cite|improve this question




















  • 2




    Please check if I've edited it correctly.
    – user376343
    Dec 7 at 21:59












  • yes, thankyou. For future, how does one use latex here?
    – Pumpkinpeach
    Dec 7 at 22:12










  • We are curious to know what the issue was with the evaluation.
    – gimusi
    Dec 7 at 22:13










  • @Pumpkinpeach Refer to MathJax basic tutorial and quick reference
    – gimusi
    Dec 7 at 22:14










  • @gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
    – Pumpkinpeach
    Dec 7 at 22:17














3












3








3







$V$ is the portion of the cone $z=sqrt{x^2+y^2};$ for $; xgeq 0.$
Find $$iiintlimits_{V} xe^{-z} dV.$$
I am trying to solve this question. The answer is supposed to be just $4.$



I have worked out the limits as



$0leq z leq infty;$ and
$;-pi/2leq theta leq pi/2;$ and $;0leq R leq z.$ What am I doing wrong?






calculus multivariable-calculus







share|cite|improve this question















$V$ is the portion of the cone $z=sqrt{x^2+y^2};$ for $; xgeq 0.$
Find $$iiintlimits_{V} xe^{-z} dV.$$
I am trying to solve this question. The answer is supposed to be just $4.$



I have worked out the limits as



$0leq z leq infty;$ and
$;-pi/2leq theta leq pi/2;$ and $;0leq R leq z.$ What am I doing wrong?






calculus multivariable-calculus




calculus multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 at 4:40









Andrews

374317




374317










asked Dec 7 at 21:44









Pumpkinpeach

628




628








  • 2




    Please check if I've edited it correctly.
    – user376343
    Dec 7 at 21:59












  • yes, thankyou. For future, how does one use latex here?
    – Pumpkinpeach
    Dec 7 at 22:12










  • We are curious to know what the issue was with the evaluation.
    – gimusi
    Dec 7 at 22:13










  • @Pumpkinpeach Refer to MathJax basic tutorial and quick reference
    – gimusi
    Dec 7 at 22:14










  • @gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
    – Pumpkinpeach
    Dec 7 at 22:17














  • 2




    Please check if I've edited it correctly.
    – user376343
    Dec 7 at 21:59












  • yes, thankyou. For future, how does one use latex here?
    – Pumpkinpeach
    Dec 7 at 22:12










  • We are curious to know what the issue was with the evaluation.
    – gimusi
    Dec 7 at 22:13










  • @Pumpkinpeach Refer to MathJax basic tutorial and quick reference
    – gimusi
    Dec 7 at 22:14










  • @gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
    – Pumpkinpeach
    Dec 7 at 22:17








2




2




Please check if I've edited it correctly.
– user376343
Dec 7 at 21:59






Please check if I've edited it correctly.
– user376343
Dec 7 at 21:59














yes, thankyou. For future, how does one use latex here?
– Pumpkinpeach
Dec 7 at 22:12




yes, thankyou. For future, how does one use latex here?
– Pumpkinpeach
Dec 7 at 22:12












We are curious to know what the issue was with the evaluation.
– gimusi
Dec 7 at 22:13




We are curious to know what the issue was with the evaluation.
– gimusi
Dec 7 at 22:13












@Pumpkinpeach Refer to MathJax basic tutorial and quick reference
– gimusi
Dec 7 at 22:14




@Pumpkinpeach Refer to MathJax basic tutorial and quick reference
– gimusi
Dec 7 at 22:14












@gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
– Pumpkinpeach
Dec 7 at 22:17




@gimusi for the calculation for integral of z^3e^-z as in @ lucky's answer below (line 5) I got z^4/4 (e^-z) and if I evaluate that between z limits I got 0?
– Pumpkinpeach
Dec 7 at 22:17










2 Answers
2






active

oldest

votes


















3














With the bounds you have stated, the integral is in fact finite:
begin{align*}
iiint_Vxe^{-z}dV&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.






share|cite|improve this answer





















  • Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
    – Pumpkinpeach
    Dec 7 at 22:14










  • Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
    – Lucky
    Dec 7 at 22:16



















2














According to the limit you have indicate, which seems to be correct, we should have



$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$



Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.



Note that the one presented here works fine: Integral evaluation.






share|cite|improve this answer

















  • 1




    $dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
    – Lucky
    Dec 7 at 22:10













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














With the bounds you have stated, the integral is in fact finite:
begin{align*}
iiint_Vxe^{-z}dV&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.






share|cite|improve this answer





















  • Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
    – Pumpkinpeach
    Dec 7 at 22:14










  • Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
    – Lucky
    Dec 7 at 22:16
















3














With the bounds you have stated, the integral is in fact finite:
begin{align*}
iiint_Vxe^{-z}dV&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.






share|cite|improve this answer





















  • Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
    – Pumpkinpeach
    Dec 7 at 22:14










  • Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
    – Lucky
    Dec 7 at 22:16














3












3








3






With the bounds you have stated, the integral is in fact finite:
begin{align*}
iiint_Vxe^{-z}dV&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.






share|cite|improve this answer












With the bounds you have stated, the integral is in fact finite:
begin{align*}
iiint_Vxe^{-z}dV&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 at 22:07









Lucky

14015




14015












  • Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
    – Pumpkinpeach
    Dec 7 at 22:14










  • Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
    – Lucky
    Dec 7 at 22:16


















  • Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
    – Pumpkinpeach
    Dec 7 at 22:14










  • Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
    – Lucky
    Dec 7 at 22:16
















Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
– Pumpkinpeach
Dec 7 at 22:14




Thanks! Can I ask how you got −e−z(z3+3z2+3z+6) in line 3? Is that just a condensed version of the by parts?
– Pumpkinpeach
Dec 7 at 22:14












Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
– Lucky
Dec 7 at 22:16




Yes. I did it by parts by hand... then checked good old Wolfram Alpha to make sure I had not made a silly mistake, which had simplified it a bit.
– Lucky
Dec 7 at 22:16











2














According to the limit you have indicate, which seems to be correct, we should have



$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$



Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.



Note that the one presented here works fine: Integral evaluation.






share|cite|improve this answer

















  • 1




    $dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
    – Lucky
    Dec 7 at 22:10


















2














According to the limit you have indicate, which seems to be correct, we should have



$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$



Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.



Note that the one presented here works fine: Integral evaluation.






share|cite|improve this answer

















  • 1




    $dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
    – Lucky
    Dec 7 at 22:10
















2












2








2






According to the limit you have indicate, which seems to be correct, we should have



$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$



Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.



Note that the one presented here works fine: Integral evaluation.






share|cite|improve this answer












According to the limit you have indicate, which seems to be correct, we should have



$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$



Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.



Note that the one presented here works fine: Integral evaluation.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 at 22:03









gimusi

1




1








  • 1




    $dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
    – Lucky
    Dec 7 at 22:10
















  • 1




    $dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
    – Lucky
    Dec 7 at 22:10










1




1




$dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
– Lucky
Dec 7 at 22:10






$dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
– Lucky
Dec 7 at 22:10




















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