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how to find the upper and lower bound for the triple integral for the tetrahedron
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i have a triple integral âˆÂzdV over the tetrahedron with vertices: (0, 0, 1), (0, 1, 1), (1, 0, 0), and (0, 0, 0).
I calculated the plane and got the boundary of 0≦x≦1, 0≦y≦1-x, 0≦z≦y.
the correct answer is supposed tobe 1/16 ,however, i keep getting the wrong answer
Can somebody help me where i get wrong and what the correct answer is?
Thank you!!!
calculus integration multivariable-calculus derivatives
asked Aug 2 at 11:56
Minto
6
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up vote
0
down vote
favorite
i have a triple integral âˆÂzdV over the tetrahedron with vertices: (0, 0, 1), (0, 1, 1), (1, 0, 0), and (0, 0, 0).
I calculated the plane and got the boundary of 0≦x≦1, 0≦y≦1-x, 0≦z≦y.
the correct answer is supposed tobe 1/16 ,however, i keep getting the wrong answer
Can somebody help me where i get wrong and what the correct answer is?
Thank you!!!
calculus integration multivariable-calculus derivatives
asked Aug 2 at 11:56
Minto
6
What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
i have a triple integral âˆÂzdV over the tetrahedron with vertices: (0, 0, 1), (0, 1, 1), (1, 0, 0), and (0, 0, 0).
I calculated the plane and got the boundary of 0≦x≦1, 0≦y≦1-x, 0≦z≦y.
the correct answer is supposed tobe 1/16 ,however, i keep getting the wrong answer
Can somebody help me where i get wrong and what the correct answer is?
Thank you!!!
calculus integration multivariable-calculus derivatives
asked Aug 2 at 11:56
Minto
6
i have a triple integral âˆÂzdV over the tetrahedron with vertices: (0, 0, 1), (0, 1, 1), (1, 0, 0), and (0, 0, 0).
I calculated the plane and got the boundary of 0≦x≦1, 0≦y≦1-x, 0≦z≦y.
the correct answer is supposed tobe 1/16 ,however, i keep getting the wrong answer
Can somebody help me where i get wrong and what the correct answer is?
Thank you!!!
calculus integration multivariable-calculus derivatives
asked Aug 2 at 11:56
Minto
6
asked Aug 2 at 11:56
Minto
6
asked Aug 2 at 11:56
Minto
6
asked Aug 2 at 11:56
Minto
6
6
What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28
add a comment |Â
What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28
What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28
What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28
add a comment |Â
1 Answer
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(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities
$$xgeq0,quad ygeq0,quad zgeq y,quad x+zleq 1 .tag1$$
Replacing the $geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $zgeq0$ in all points of $T$. We therefore choose the triangle $T':=bigl(x,0,z)bigm$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)in T$ one necessarily has $(x,0,z)in T'$. Given a point $(x,0,z)in T'$ we now have to determine the set of $y$ for which $(x,y,z)in T$. Looking at $(1)$ we see that this is the $y$-interval $0leq yleq z$. We therefore obtain
$$eqalignint_T z>rm d(x,y,z)&=int_T' zleft(int_0^z dyright)>rm d(x,z)=int_T'z^2> rm d(x,z)cr &=int_0^1 z^2left(int_0^1-z dxright)dz=int_0^1(z^2-z^3)>dzcr &=1over12 .cr $$
answered Aug 2 at 14:37
Christian Blatter
163k7106305
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities
$$xgeq0,quad ygeq0,quad zgeq y,quad x+zleq 1 .tag1$$
Replacing the $geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $zgeq0$ in all points of $T$. We therefore choose the triangle $T':=bigl(x,0,z)bigm$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)in T$ one necessarily has $(x,0,z)in T'$. Given a point $(x,0,z)in T'$ we now have to determine the set of $y$ for which $(x,y,z)in T$. Looking at $(1)$ we see that this is the $y$-interval $0leq yleq z$. We therefore obtain
$$eqalignint_T z>rm d(x,y,z)&=int_T' zleft(int_0^z dyright)>rm d(x,z)=int_T'z^2> rm d(x,z)cr &=int_0^1 z^2left(int_0^1-z dxright)dz=int_0^1(z^2-z^3)>dzcr &=1over12 .cr $$
answered Aug 2 at 14:37
Christian Blatter
163k7106305
add a comment |Â
up vote
0
down vote
(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities
$$xgeq0,quad ygeq0,quad zgeq y,quad x+zleq 1 .tag1$$
Replacing the $geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $zgeq0$ in all points of $T$. We therefore choose the triangle $T':=bigl(x,0,z)bigm$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)in T$ one necessarily has $(x,0,z)in T'$. Given a point $(x,0,z)in T'$ we now have to determine the set of $y$ for which $(x,y,z)in T$. Looking at $(1)$ we see that this is the $y$-interval $0leq yleq z$. We therefore obtain
$$eqalignint_T z>rm d(x,y,z)&=int_T' zleft(int_0^z dyright)>rm d(x,z)=int_T'z^2> rm d(x,z)cr &=int_0^1 z^2left(int_0^1-z dxright)dz=int_0^1(z^2-z^3)>dzcr &=1over12 .cr $$
answered Aug 2 at 14:37
Christian Blatter
163k7106305
add a comment |Â
up vote
0
down vote
up vote
0
down vote
(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities
$$xgeq0,quad ygeq0,quad zgeq y,quad x+zleq 1 .tag1$$
Replacing the $geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $zgeq0$ in all points of $T$. We therefore choose the triangle $T':=bigl(x,0,z)bigm$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)in T$ one necessarily has $(x,0,z)in T'$. Given a point $(x,0,z)in T'$ we now have to determine the set of $y$ for which $(x,y,z)in T$. Looking at $(1)$ we see that this is the $y$-interval $0leq yleq z$. We therefore obtain
$$eqalignint_T z>rm d(x,y,z)&=int_T' zleft(int_0^z dyright)>rm d(x,z)=int_T'z^2> rm d(x,z)cr &=int_0^1 z^2left(int_0^1-z dxright)dz=int_0^1(z^2-z^3)>dzcr &=1over12 .cr $$
answered Aug 2 at 14:37
Christian Blatter
163k7106305
(Your planes do not correspond to the given points.)
Draw a figure! You then will see that the tetrahedron $T$ is defined by the four inequalities
$$xgeq0,quad ygeq0,quad zgeq y,quad x+zleq 1 .tag1$$
Replacing the $geq$ signs with $=$ we obtain the equations of the four planes containing the faces of $T$. Any three of these four planes intersect in one of the given vertices.
From $(1)$ we infer that automatically $zgeq0$ in all points of $T$. We therefore choose the triangle $T':=bigl(x,0,z)bigm$ as domain for the two-dimensional outer integral. Note that for any point $(x,y,z)in T$ one necessarily has $(x,0,z)in T'$. Given a point $(x,0,z)in T'$ we now have to determine the set of $y$ for which $(x,y,z)in T$. Looking at $(1)$ we see that this is the $y$-interval $0leq yleq z$. We therefore obtain
$$eqalignint_T z>rm d(x,y,z)&=int_T' zleft(int_0^z dyright)>rm d(x,z)=int_T'z^2> rm d(x,z)cr &=int_0^1 z^2left(int_0^1-z dxright)dz=int_0^1(z^2-z^3)>dzcr &=1over12 .cr $$
answered Aug 2 at 14:37
Christian Blatter
163k7106305
answered Aug 2 at 14:37
Christian Blatter
163k7106305
answered Aug 2 at 14:37
Christian Blatter
163k7106305
answered Aug 2 at 14:37
Christian Blatter
163k7106305
163k7106305
add a comment |Â
add a comment |Â
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What is the domain over which you are integrating ?
– Kuifje
Aug 2 at 13:28