How to obtain the surface area of a coin generated by rotating a region about the $x$-axis?
$begingroup$
If we rotate the region below ( the picture shown below ) about the $x$-axis, we obtain a coin, and I want to calculate the surface area of this coin, using
$$SA=2pi int_{a}^{b} f(x)sqrt{1+f'(x)^2}$$
you can see that there are two semicircles, the upper semicircle is given by $y=2+sqrt{1-x^{2}}$, the one at the bottom is given by $y=-2-sqrt{1-x^{2}}$, what I don't understand is, in the solution manual, only the upper circle was taken into consideration.
I am confused, because in a different problem, I had a torus obtained by rotating a circle around the the $x$-axis, and both the upper half and the bottom half of circle was taken into consideration when doing the calculations... I don't understand however why in the case of a coin this is different? Why are we only taking the upper semicircle? What am I missing here exactly? I am not asking for a solution, but rather an explanation as to why.
calculus
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
$endgroup$
add a comment |
$begingroup$
If we rotate the region below ( the picture shown below ) about the $x$-axis, we obtain a coin, and I want to calculate the surface area of this coin, using
$$SA=2pi int_{a}^{b} f(x)sqrt{1+f'(x)^2}$$
you can see that there are two semicircles, the upper semicircle is given by $y=2+sqrt{1-x^{2}}$, the one at the bottom is given by $y=-2-sqrt{1-x^{2}}$, what I don't understand is, in the solution manual, only the upper circle was taken into consideration.
I am confused, because in a different problem, I had a torus obtained by rotating a circle around the the $x$-axis, and both the upper half and the bottom half of circle was taken into consideration when doing the calculations... I don't understand however why in the case of a coin this is different? Why are we only taking the upper semicircle? What am I missing here exactly? I am not asking for a solution, but rather an explanation as to why.
calculus
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
$endgroup$
add a comment |
$begingroup$
If we rotate the region below ( the picture shown below ) about the $x$-axis, we obtain a coin, and I want to calculate the surface area of this coin, using
$$SA=2pi int_{a}^{b} f(x)sqrt{1+f'(x)^2}$$
you can see that there are two semicircles, the upper semicircle is given by $y=2+sqrt{1-x^{2}}$, the one at the bottom is given by $y=-2-sqrt{1-x^{2}}$, what I don't understand is, in the solution manual, only the upper circle was taken into consideration.
I am confused, because in a different problem, I had a torus obtained by rotating a circle around the the $x$-axis, and both the upper half and the bottom half of circle was taken into consideration when doing the calculations... I don't understand however why in the case of a coin this is different? Why are we only taking the upper semicircle? What am I missing here exactly? I am not asking for a solution, but rather an explanation as to why.
calculus
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
$endgroup$
If we rotate the region below ( the picture shown below ) about the $x$-axis, we obtain a coin, and I want to calculate the surface area of this coin, using
$$SA=2pi int_{a}^{b} f(x)sqrt{1+f'(x)^2}$$
you can see that there are two semicircles, the upper semicircle is given by $y=2+sqrt{1-x^{2}}$, the one at the bottom is given by $y=-2-sqrt{1-x^{2}}$, what I don't understand is, in the solution manual, only the upper circle was taken into consideration.
I am confused, because in a different problem, I had a torus obtained by rotating a circle around the the $x$-axis, and both the upper half and the bottom half of circle was taken into consideration when doing the calculations... I don't understand however why in the case of a coin this is different? Why are we only taking the upper semicircle? What am I missing here exactly? I am not asking for a solution, but rather an explanation as to why.
calculus
calculus
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
asked Dec 25 '18 at 11:14
D. QaD. Qa
1656
1656
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
$endgroup$
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
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%2f3052016%2fhow-to-obtain-the-surface-area-of-a-coin-generated-by-rotating-a-region-about-th%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
$endgroup$
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
add a comment |
$begingroup$
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
$endgroup$
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
add a comment |
$begingroup$
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
$endgroup$
The cross section of the coin in the $xy$ plane is symmetric around the $x$ axis. Since one rotates by $2pi$ around the $x$ axis, if both top and bottom were used, then it would sweep through the coin twice. So by only using top part it sweeps only once as desired.
In the related torus problem, was it the case that the entire circle rotated was all on one side of the rotation axis? If so that would explain using both top and bottom half of the circle.
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
edited Dec 26 '18 at 0:02
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
answered Dec 25 '18 at 11:20
coffeemathcoffeemath
2,9171415
2,9171415
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
add a comment |
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
1
1
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
$begingroup$
I understood now, for the tours problem the entire circle was rotated so that explains why we took the top and bottom semicircle into consideration. Thank you, that cleared my doubts :)
$endgroup$
– D. Qa
Dec 26 '18 at 14:33
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.
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%2f3052016%2fhow-to-obtain-the-surface-area-of-a-coin-generated-by-rotating-a-region-about-th%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
