Mixing
give 5 other equivalent iterated triple integrals
$begingroup$
I am given the following integral:
$$int_0^2int_0^{y^3}int_0^{y^2}f(x,y,z) dzdxdy $$
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$int_0^8int_0^{x^{2/3}}int_{x^{1/3}}^2f(x,y,z)dydzdx,+,int_0^8int_{x^{2/3}}^4int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split?
calculus integration multivariable-calculus
asked Jan 28 at 23:49
user10939145user10939145
111
$endgroup$
add a comment |
$begingroup$
I am given the following integral:
$$int_0^2int_0^{y^3}int_0^{y^2}f(x,y,z) dzdxdy $$
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$int_0^8int_0^{x^{2/3}}int_{x^{1/3}}^2f(x,y,z)dydzdx,+,int_0^8int_{x^{2/3}}^4int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split?
calculus integration multivariable-calculus
asked Jan 28 at 23:49
user10939145user10939145
111
$endgroup$
add a comment |
$begingroup$
I am given the following integral:
$$int_0^2int_0^{y^3}int_0^{y^2}f(x,y,z) dzdxdy $$
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$int_0^8int_0^{x^{2/3}}int_{x^{1/3}}^2f(x,y,z)dydzdx,+,int_0^8int_{x^{2/3}}^4int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split?
calculus integration multivariable-calculus
asked Jan 28 at 23:49
user10939145user10939145
111
$endgroup$
I am given the following integral:
$$int_0^2int_0^{y^3}int_0^{y^2}f(x,y,z) dzdxdy $$
I was successfully able to rewrite this in its dzdydx, dxdzdy, and dxdydz forms, but I'm having a hard time understanding the iterated triple integrals where dy is the inner integral.
The solution for dydzdx is as follows:
$$int_0^8int_0^{x^{2/3}}int_{x^{1/3}}^2f(x,y,z)dydzdx,+,int_0^8int_{x^{2/3}}^4int_{z^{1/2}}^2f(x,y,z)dydzdx $$
I don't understand why the integral was split. How do I figure out what this shape looks like in the zx plane in order to even know that it needed to be split?
calculus integration multivariable-calculus
calculus integration multivariable-calculus
asked Jan 28 at 23:49
user10939145user10939145
111
asked Jan 28 at 23:49
user10939145user10939145
111
asked Jan 28 at 23:49
user10939145user10939145
111
asked Jan 28 at 23:49
user10939145user10939145
111
asked Jan 28 at 23:49
user10939145user10939145
111
111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done.
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
add a comment |
$begingroup$
The region of the integral is characterized by
$$
0 leq y leq 2
\ 0 leq x leq y^3 \ 0 leq z leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 leq x leq y^3 leq 2^3 = 8 \ 0 leq z leq y^2 leq 4$$
The $0 leq x leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y geq x^{1/3} \ y geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $zleq x^{2/3}$ and the limits will be
$$
int_{x=0}^8int_{z= 0}^{x^{2/3}} int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z leq y^2 leq 4$ and the limits will be
$$
int_{x=0}^8int_{z= x^{2/3}}^{4} int_{y=z^{1/2}}^2
$$
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
$endgroup$
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
$begingroup$
The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done.
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
add a comment |
$begingroup$
The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done.
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
add a comment |
$begingroup$
The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done.
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
$endgroup$
The conditions on $y$ are $y^{2} >z$ and $y^{3} >x$. We have to integrate w.r.t. $y$ from the maximum of $sqrt z$ and $x^{1/3}$ to $2$. In order to carry this out it is convenient to split the possible values of $x$ and $z$ into two parts: $x>z^{2/3}$ and $x<z^{2/3}$ so that we know what the maximum is. This is what they have done.
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
answered Jan 29 at 0:03
Kavi Rama MurthyKavi Rama Murthy
71.1k53170
71.1k53170
add a comment |
add a comment |
$begingroup$
The region of the integral is characterized by
$$
0 leq y leq 2
\ 0 leq x leq y^3 \ 0 leq z leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 leq x leq y^3 leq 2^3 = 8 \ 0 leq z leq y^2 leq 4$$
The $0 leq x leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y geq x^{1/3} \ y geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $zleq x^{2/3}$ and the limits will be
$$
int_{x=0}^8int_{z= 0}^{x^{2/3}} int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z leq y^2 leq 4$ and the limits will be
$$
int_{x=0}^8int_{z= x^{2/3}}^{4} int_{y=z^{1/2}}^2
$$
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
$endgroup$
add a comment |
$begingroup$
The region of the integral is characterized by
$$
0 leq y leq 2
\ 0 leq x leq y^3 \ 0 leq z leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 leq x leq y^3 leq 2^3 = 8 \ 0 leq z leq y^2 leq 4$$
The $0 leq x leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y geq x^{1/3} \ y geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $zleq x^{2/3}$ and the limits will be
$$
int_{x=0}^8int_{z= 0}^{x^{2/3}} int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z leq y^2 leq 4$ and the limits will be
$$
int_{x=0}^8int_{z= x^{2/3}}^{4} int_{y=z^{1/2}}^2
$$
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
$endgroup$
add a comment |
$begingroup$
The region of the integral is characterized by
$$
0 leq y leq 2
\ 0 leq x leq y^3 \ 0 leq z leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 leq x leq y^3 leq 2^3 = 8 \ 0 leq z leq y^2 leq 4$$
The $0 leq x leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y geq x^{1/3} \ y geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $zleq x^{2/3}$ and the limits will be
$$
int_{x=0}^8int_{z= 0}^{x^{2/3}} int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z leq y^2 leq 4$ and the limits will be
$$
int_{x=0}^8int_{z= x^{2/3}}^{4} int_{y=z^{1/2}}^2
$$
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
$endgroup$
The region of the integral is characterized by
$$
0 leq y leq 2
\ 0 leq x leq y^3 \ 0 leq z leq y^2
$$
If we have in mind that $x$ should be the outermost variable, the first thing is to deduce that $$0 leq x leq y^3 leq 2^3 = 8 \ 0 leq z leq y^2 leq 4$$
The $0 leq x leq 2^3 = 8$ part gives the outer integral limits.
And we also have two lower-boundary conditions on $y$, namely
$$ y geq x^{1/3} \ y geq z^{1/2}$$
And here is why the integral has to be split up -- we have no clue whether $x^{1/3}$ is more or less than $z^{1/2}$. If it is greater or equal, then $zleq x^{2/3}$ and the limits will be
$$
int_{x=0}^8int_{z= 0}^{x^{2/3}} int_{y=x^{1/3}}^2
$$
and if $x^{1/3} < z^{1/2}$ the upper limit on $z$ is given by $z leq y^2 leq 4$ and the limits will be
$$
int_{x=0}^8int_{z= x^{2/3}}^{4} int_{y=z^{1/2}}^2
$$
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
answered Jan 29 at 0:22
Mark FischlerMark Fischler
33.8k12552
33.8k12552
add a comment |
add a comment |
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