Can we calculate double integrals without drawing the sketch?

Multi tool use
Multi tool use












0












$begingroup$


I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?



For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    why you downvote it @Matti P. ??
    $endgroup$
    – Hamza.S
    Jan 22 at 8:21












  • $begingroup$
    I didn't, it was someone else
    $endgroup$
    – Matti P.
    Jan 22 at 8:22






  • 1




    $begingroup$
    I am not the downvoter, but to prevent more downvotes you should show an attempt.
    $endgroup$
    – Robert Z
    Jan 22 at 8:24








  • 4




    $begingroup$
    Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
    $endgroup$
    – Shubham Johri
    Jan 22 at 8:25






  • 4




    $begingroup$
    But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 8:34


















0












$begingroup$


I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?



For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.






multivariable-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    why you downvote it @Matti P. ??
    $endgroup$
    – Hamza.S
    Jan 22 at 8:21












  • $begingroup$
    I didn't, it was someone else
    $endgroup$
    – Matti P.
    Jan 22 at 8:22






  • 1




    $begingroup$
    I am not the downvoter, but to prevent more downvotes you should show an attempt.
    $endgroup$
    – Robert Z
    Jan 22 at 8:24








  • 4




    $begingroup$
    Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
    $endgroup$
    – Shubham Johri
    Jan 22 at 8:25






  • 4




    $begingroup$
    But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 8:34
















0












0








0





$begingroup$


I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?



For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.






multivariable-calculus







share|cite|improve this question











$endgroup$




I found it very hard to draw the sketch so my question is can we solve the question numerically without drawing the sketch ?



For example I have a question
$$iint_D arctan{frac{y}{x}},dxdy$$
where $D$ is domain bounded by $x^2+y^2=1$, $x^2+y^2=4$, $y=x$, $y = sqrt{3}x$.






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 8:54









Jyrki Lahtonen

109k13170376




109k13170376










asked Jan 22 at 8:19









Hamza.SHamza.S

1236




1236












  • $begingroup$
    why you downvote it @Matti P. ??
    $endgroup$
    – Hamza.S
    Jan 22 at 8:21












  • $begingroup$
    I didn't, it was someone else
    $endgroup$
    – Matti P.
    Jan 22 at 8:22






  • 1




    $begingroup$
    I am not the downvoter, but to prevent more downvotes you should show an attempt.
    $endgroup$
    – Robert Z
    Jan 22 at 8:24








  • 4




    $begingroup$
    Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
    $endgroup$
    – Shubham Johri
    Jan 22 at 8:25






  • 4




    $begingroup$
    But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 8:34




















  • $begingroup$
    why you downvote it @Matti P. ??
    $endgroup$
    – Hamza.S
    Jan 22 at 8:21












  • $begingroup$
    I didn't, it was someone else
    $endgroup$
    – Matti P.
    Jan 22 at 8:22






  • 1




    $begingroup$
    I am not the downvoter, but to prevent more downvotes you should show an attempt.
    $endgroup$
    – Robert Z
    Jan 22 at 8:24








  • 4




    $begingroup$
    Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
    $endgroup$
    – Shubham Johri
    Jan 22 at 8:25






  • 4




    $begingroup$
    But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
    $endgroup$
    – Jyrki Lahtonen
    Jan 22 at 8:34


















$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21






$begingroup$
why you downvote it @Matti P. ??
$endgroup$
– Hamza.S
Jan 22 at 8:21














$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22




$begingroup$
I didn't, it was someone else
$endgroup$
– Matti P.
Jan 22 at 8:22




1




1




$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24






$begingroup$
I am not the downvoter, but to prevent more downvotes you should show an attempt.
$endgroup$
– Robert Z
Jan 22 at 8:24






4




4




$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25




$begingroup$
Solving a double integral when the limits are explicitly given is easy. Solving it on a domain requires figuring out the limits that is very difficult to do without a physical or mental sketch of the domain. In your case, however, you can simplify the integral and the domain by using polar coordinates
$endgroup$
– Shubham Johri
Jan 22 at 8:25




4




4




$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34






$begingroup$
But, yes, you definitely should use the polar coordinates here. And, yes, I believe in drawing pictures in cases like these. The reason is that in applications, particularly those originating in physics, the boundaries are defined visually rather than using equations. If you cannot figure out the boundaries from a given image, you absolutely must practice that process.
$endgroup$
– Jyrki Lahtonen
Jan 22 at 8:34












1 Answer
1






active

oldest

votes


















2












$begingroup$

You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.



Keep $x^2+y^2=r^2,theta=arctan(y/x)$.



The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.



Figure



The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
    $endgroup$
    – Hamza.S
    Jan 22 at 8:38






  • 1




    $begingroup$
    @Hamas.S If you are not familiar with graphing polar curves, look here.
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:00










  • $begingroup$
    Ok Thanks @ShubhamJohri
    $endgroup$
    – Hamza.S
    Jan 22 at 9:02






  • 2




    $begingroup$
    @Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:05













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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082881%2fcan-we-calculate-double-integrals-without-drawing-the-sketch%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









2












$begingroup$

You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.



Keep $x^2+y^2=r^2,theta=arctan(y/x)$.



The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.



Figure



The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
    $endgroup$
    – Hamza.S
    Jan 22 at 8:38






  • 1




    $begingroup$
    @Hamas.S If you are not familiar with graphing polar curves, look here.
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:00










  • $begingroup$
    Ok Thanks @ShubhamJohri
    $endgroup$
    – Hamza.S
    Jan 22 at 9:02






  • 2




    $begingroup$
    @Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:05


















2












$begingroup$

You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.



Keep $x^2+y^2=r^2,theta=arctan(y/x)$.



The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.



Figure



The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
    $endgroup$
    – Hamza.S
    Jan 22 at 8:38






  • 1




    $begingroup$
    @Hamas.S If you are not familiar with graphing polar curves, look here.
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:00










  • $begingroup$
    Ok Thanks @ShubhamJohri
    $endgroup$
    – Hamza.S
    Jan 22 at 9:02






  • 2




    $begingroup$
    @Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:05
















2












2








2





$begingroup$

You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.



Keep $x^2+y^2=r^2,theta=arctan(y/x)$.



The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.



Figure



The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$






share|cite|improve this answer











$endgroup$



You will eventually have to make a physical or mental sketch of the domain to figure out the limits of integration. You can try to simplify the given equations through some kind of substitution. Polar coordinates work well here.



Keep $x^2+y^2=r^2,theta=arctan(y/x)$.



The domain $D$ is flanked by $r=1,r=2,theta=pi/4,theta=pi/3$. This is pretty simple to sketch.



Figure



The integral in the first quadrant becomes$$int_{pi/4}^{pi/3}int_1^2theta rdr~dtheta$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 22 at 9:13

























answered Jan 22 at 8:36









Shubham JohriShubham Johri

5,189718




5,189718












  • $begingroup$
    thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
    $endgroup$
    – Hamza.S
    Jan 22 at 8:38






  • 1




    $begingroup$
    @Hamas.S If you are not familiar with graphing polar curves, look here.
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:00










  • $begingroup$
    Ok Thanks @ShubhamJohri
    $endgroup$
    – Hamza.S
    Jan 22 at 9:02






  • 2




    $begingroup$
    @Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:05




















  • $begingroup$
    thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
    $endgroup$
    – Hamza.S
    Jan 22 at 8:38






  • 1




    $begingroup$
    @Hamas.S If you are not familiar with graphing polar curves, look here.
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:00










  • $begingroup$
    Ok Thanks @ShubhamJohri
    $endgroup$
    – Hamza.S
    Jan 22 at 9:02






  • 2




    $begingroup$
    @Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
    $endgroup$
    – Shubham Johri
    Jan 22 at 9:05


















$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38




$begingroup$
thanks . but sorry for the stupid question i am a beginner i found the sketching very hard to draw .
$endgroup$
– Hamza.S
Jan 22 at 8:38




1




1




$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00




$begingroup$
@Hamas.S If you are not familiar with graphing polar curves, look here.
$endgroup$
– Shubham Johri
Jan 22 at 9:00












$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02




$begingroup$
Ok Thanks @ShubhamJohri
$endgroup$
– Hamza.S
Jan 22 at 9:02




2




2




$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05






$begingroup$
@Hamza.S Note that I used polar coordinates for ease of integration. The integrand $arctan(y/x)$ is fairly difficult to integrate in Cartesian coordinates. In polar coordinates, it reduces to a mere $theta$ in the first quadrant. For the purpose of drawing the sketch, it is equally easy to observe that in the Cartesian system, the first two equations represent circles centred at $(0,0)$ with radii $1,2$ respectively while the the latter represent straight lines passing through $(0,0)$ and inclined at $45deg,60deg$ in the anticlockwise direction with the positive $x$ axis respectively
$endgroup$
– Shubham Johri
Jan 22 at 9:05




















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082881%2fcan-we-calculate-double-integrals-without-drawing-the-sketch%23new-answer', 'question_page');
}
);

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







2wWGZgjdP2JrYcSkjZxPBzi mk9ik,QO6Dmj7Ipr1RpTWeQIbJp7wWHh8VjDi aI2RA9Ny8O,p4fe
-
bd5dBuwCVUhDjpsv,s6X CH wL7wU6E2pSjz0h

Popular posts from this blog

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

menu

Image
menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
Read more

Dobbiaco

Image
Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
Read more