Triple integral of portion of cone (cylindrical polar coordinates)?
up vote
3
down vote
favorite
$V$ 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
edited Dec 8 at 4:40
Andrews
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
|
show 3 more comments
up vote
3
down vote
favorite
$V$ 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
edited Dec 8 at 4:40
Andrews
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
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
|
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$V$ 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
edited Dec 8 at 4:40
Andrews
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
$V$ 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
edited Dec 8 at 4:40
Andrews
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
edited Dec 8 at 4:40
Andrews
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
edited Dec 8 at 4:40
Andrews
322217
edited Dec 8 at 4:40
Andrews
322217
edited Dec 8 at 4:40
Andrews
322217
322217
asked Dec 7 at 21:44
Pumpkinpeach
538
asked Dec 7 at 21:44
Pumpkinpeach
538
asked Dec 7 at 21:44
Pumpkinpeach
538
538
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
|
show 3 more comments
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
|
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
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.
answered Dec 7 at 22:07
Lucky
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
add a comment |
up vote
2
down vote
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.
answered Dec 7 at 22:03
gimusi
1
1
$dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
– Lucky
Dec 7 at 22:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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.
answered Dec 7 at 22:07
Lucky
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
add a comment |
up vote
3
down vote
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.
answered Dec 7 at 22:07
Lucky
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
add a comment |
up vote
3
down vote
up vote
3
down vote
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.
answered Dec 7 at 22:07
Lucky
14015
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.
answered Dec 7 at 22:07
Lucky
14015
answered Dec 7 at 22:07
Lucky
14015
answered Dec 7 at 22:07
Lucky
14015
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
add a comment |
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
add a comment |
up vote
2
down vote
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.
answered Dec 7 at 22:03
gimusi
1
1
$dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
– Lucky
Dec 7 at 22:10
add a comment |
up vote
2
down vote
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.
answered Dec 7 at 22:03
gimusi
1
1
$dV=rdrdzdtheta$ is correct. The Jacobian is just $r$.
– Lucky
Dec 7 at 22:10
add a comment |
up vote
2
down vote
up vote
2
down vote
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.
answered Dec 7 at 22:03
gimusi
1
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.
answered Dec 7 at 22:03
gimusi
1
answered Dec 7 at 22:03
gimusi
1
answered Dec 7 at 22:03
gimusi
1
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
add a comment |
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
add a comment |
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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