Use only double integrals to find the volume of a solid tetrahedron
Use only double integrals to find the volume of the solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,2,0)$ and $(2,2,0)$.
I know you plot the points on $xyz$-plane, but how do you get the equation of the plane that goes in the equation by the four vertices?
multivariable-calculus
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
add a comment |
Use only double integrals to find the volume of the solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,2,0)$ and $(2,2,0)$.
I know you plot the points on $xyz$-plane, but how do you get the equation of the plane that goes in the equation by the four vertices?
multivariable-calculus
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
1
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55
add a comment |
Use only double integrals to find the volume of the solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,2,0)$ and $(2,2,0)$.
I know you plot the points on $xyz$-plane, but how do you get the equation of the plane that goes in the equation by the four vertices?
multivariable-calculus
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
Use only double integrals to find the volume of the solid tetrahedron with vertices $(0,0,0)$, $(0,0,1)$, $(0,2,0)$ and $(2,2,0)$.
I know you plot the points on $xyz$-plane, but how do you get the equation of the plane that goes in the equation by the four vertices?
multivariable-calculus
multivariable-calculus
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
edited Aug 28 '17 at 1:01
Chase Ryan Taylor
4,38021530
4,38021530
asked Oct 3 '14 at 21:27
user177780
162
asked Oct 3 '14 at 21:27
user177780
162
asked Oct 3 '14 at 21:27
user177780
162
162
1
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55
add a comment |
1
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55
1
1
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55
add a comment |
2 Answers
2
active
oldest
votes
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)times (2,2,-1)$ is a normal vector for the plane you want.
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
add a comment |
Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.
answered Jan 23 '18 at 18:52
G Cab
18k31237
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f957313%2fuse-only-double-integrals-to-find-the-volume-of-a-solid-tetrahedron%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)times (2,2,-1)$ is a normal vector for the plane you want.
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
add a comment |
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)times (2,2,-1)$ is a normal vector for the plane you want.
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
add a comment |
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)times (2,2,-1)$ is a normal vector for the plane you want.
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
HINT: If you are going to integrate with respect to $x$ and $y$, then the vectors $(0,2,0)-(0,0,1)=(0,2,-1)$ and $(2,2,0)-(0,0,1)=(2,2,-1)$ are in the plane you want. So $(0,2,-1)times (2,2,-1)$ is a normal vector for the plane you want.
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
answered Oct 4 '14 at 1:24
Laars Helenius
6,2611323
6,2611323
add a comment |
add a comment |
Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.
answered Jan 23 '18 at 18:52
G Cab
18k31237
add a comment |
Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.
answered Jan 23 '18 at 18:52
G Cab
18k31237
add a comment |
Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.
answered Jan 23 '18 at 18:52
G Cab
18k31237
Integrating with respect to $x$ and $y$, then
- the "base" triangle in that plane is given by $(0,0,0)$,$(0,2,0)$,$(2,2,0)$, and $(x,y)$ shall vary in that triangle;
- the "roof" plane, that is the plane that bounds the $z$ height, will be the plane passing through $(0,0,1)$,$(0,2,0)$,$(2,2,0)$.
answered Jan 23 '18 at 18:52
G Cab
18k31237
answered Jan 23 '18 at 18:52
G Cab
18k31237
answered Jan 23 '18 at 18:52
G Cab
18k31237
answered Jan 23 '18 at 18:52
G Cab
18k31237
18k31237
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f957313%2fuse-only-double-integrals-to-find-the-volume-of-a-solid-tetrahedron%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
This doesn't really require calculus. The volume of the tetrahedron can be found by evaluating a determinant.
– Alan
Oct 4 '14 at 0:55
I don't understand your final sentence. Could you revise it? (Grammatically speaking, it's not coherent.) Also, there's no such thing as an "$xyz$-plane"; a plane is two-dimensional, but $x$, $y$ and $z$ are three dimensions.
– Chase Ryan Taylor
Aug 28 '17 at 0:55