Volume of hyperbola revolved about the y -axis












5












$begingroup$


I'm trying to calculate the volume of the solid formed by revolving the hyperbola ${x^2} - {y^2} = 1$ bounded by $x=1$ and $x=3$ about the y axis, however I do not know if I'm going about this the right way using cylindrical shells.



Using volume of a solid of revolution with cylindrical-shell method where the radius is ${x}$ and the height is ${2sqrt{x^2 - 1}}$, I got the integral:
$$
begin{eqnarray}
V &=& 2 pi int_1^{3} [x (2sqrt{x^2 - 1})] , textrm{d}x \
&=& 4 pi left[ frac{(x^2 - 1)^{3/2}}{3} right]_1^{3} \
&=& frac{32sqrt{8} pi}{3} \
end{eqnarray}
$$



I would like to know if this is the correct way to solve this problem using cylindrical shells and if there are any other ways to solve the this problem.






calculus integration volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you have bounds on the $x$ values?
    $endgroup$
    – Andres Mejia
    9 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    9 hours ago
















5












$begingroup$


I'm trying to calculate the volume of the solid formed by revolving the hyperbola ${x^2} - {y^2} = 1$ bounded by $x=1$ and $x=3$ about the y axis, however I do not know if I'm going about this the right way using cylindrical shells.



Using volume of a solid of revolution with cylindrical-shell method where the radius is ${x}$ and the height is ${2sqrt{x^2 - 1}}$, I got the integral:
$$
begin{eqnarray}
V &=& 2 pi int_1^{3} [x (2sqrt{x^2 - 1})] , textrm{d}x \
&=& 4 pi left[ frac{(x^2 - 1)^{3/2}}{3} right]_1^{3} \
&=& frac{32sqrt{8} pi}{3} \
end{eqnarray}
$$



I would like to know if this is the correct way to solve this problem using cylindrical shells and if there are any other ways to solve the this problem.






calculus integration volume







share|cite|improve this question











$endgroup$












  • $begingroup$
    do you have bounds on the $x$ values?
    $endgroup$
    – Andres Mejia
    9 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    9 hours ago














5












5








5





$begingroup$


I'm trying to calculate the volume of the solid formed by revolving the hyperbola ${x^2} - {y^2} = 1$ bounded by $x=1$ and $x=3$ about the y axis, however I do not know if I'm going about this the right way using cylindrical shells.



Using volume of a solid of revolution with cylindrical-shell method where the radius is ${x}$ and the height is ${2sqrt{x^2 - 1}}$, I got the integral:
$$
begin{eqnarray}
V &=& 2 pi int_1^{3} [x (2sqrt{x^2 - 1})] , textrm{d}x \
&=& 4 pi left[ frac{(x^2 - 1)^{3/2}}{3} right]_1^{3} \
&=& frac{32sqrt{8} pi}{3} \
end{eqnarray}
$$



I would like to know if this is the correct way to solve this problem using cylindrical shells and if there are any other ways to solve the this problem.






calculus integration volume







share|cite|improve this question











$endgroup$




I'm trying to calculate the volume of the solid formed by revolving the hyperbola ${x^2} - {y^2} = 1$ bounded by $x=1$ and $x=3$ about the y axis, however I do not know if I'm going about this the right way using cylindrical shells.



Using volume of a solid of revolution with cylindrical-shell method where the radius is ${x}$ and the height is ${2sqrt{x^2 - 1}}$, I got the integral:
$$
begin{eqnarray}
V &=& 2 pi int_1^{3} [x (2sqrt{x^2 - 1})] , textrm{d}x \
&=& 4 pi left[ frac{(x^2 - 1)^{3/2}}{3} right]_1^{3} \
&=& frac{32sqrt{8} pi}{3} \
end{eqnarray}
$$



I would like to know if this is the correct way to solve this problem using cylindrical shells and if there are any other ways to solve the this problem.






calculus integration volume




calculus integration volume






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 hours ago







Ludwig

















asked 9 hours ago









LudwigLudwig

805




805












  • $begingroup$
    do you have bounds on the $x$ values?
    $endgroup$
    – Andres Mejia
    9 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    9 hours ago


















  • $begingroup$
    do you have bounds on the $x$ values?
    $endgroup$
    – Andres Mejia
    9 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    9 hours ago
















$begingroup$
do you have bounds on the $x$ values?
$endgroup$
– Andres Mejia
9 hours ago




$begingroup$
do you have bounds on the $x$ values?
$endgroup$
– Andres Mejia
9 hours ago












$begingroup$
Yes, sorry, the bounds on the x values are 1 and 3
$endgroup$
– Ludwig
9 hours ago




$begingroup$
Yes, sorry, the bounds on the x values are 1 and 3
$endgroup$
– Ludwig
9 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Your solution is correct.



Method 2: Using double integrals.



Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=sqrt{x^2+z^2-1}$, for $yge 0$. Next, define a region



$$D={(x,z) | 1le x^2+z^2 le 9}$$



To get the volume of the upper body, we evaluate the integral



$$iintlimits_D y(x,z) text dx text dz = iintlimits_D sqrt{x^2+z^2-1} text dx text dz$$



and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:



$$V = 2int_0^{2pi}int_1^3 rsqrt{r^2-1} text dr text dtheta$$



Method 3: The washer method.



Consider a horizontal washer (ring) with a thickness of $text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $text dV = (r_2^2-r_1^2)pi$. To get the total volume, integrate the volumes of all such washers:



$$V=intlimits_{-2sqrt2}^{2sqrt2} pi(9-y^2-1) text dy$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
    $endgroup$
    – Ludwig
    8 hours ago










  • $begingroup$
    @Ludwig I apologize, I have misread your question. I have updated my answer.
    $endgroup$
    – Haris Gusic
    8 hours ago



















4












$begingroup$

The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.



$$
V=2cdot 2piint_{1}^{3}xsqrt{x^2-1},dx=
frac{4}{2}piint_{1}^{3}sqrt{x^2-1}frac{d}{dx}(x^2-1),dx=\
2piint_{0}^{8}sqrt{u},du=2pifrac{2sqrt{u^3}}{3}bigg|_{0}^{8}=
frac{4}{3}pileft(sqrt{8^3}-sqrt{0}right)=frac{64sqrt{2}pi}{3}
$$



Wolfram Alpha check






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    8 hours ago






  • 1




    $begingroup$
    I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
    $endgroup$
    – Michael Rybkin
    8 hours ago













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%2f3141433%2fvolume-of-hyperbola-revolved-about-the-y-axis%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









3












$begingroup$

Your solution is correct.



Method 2: Using double integrals.



Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=sqrt{x^2+z^2-1}$, for $yge 0$. Next, define a region



$$D={(x,z) | 1le x^2+z^2 le 9}$$



To get the volume of the upper body, we evaluate the integral



$$iintlimits_D y(x,z) text dx text dz = iintlimits_D sqrt{x^2+z^2-1} text dx text dz$$



and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:



$$V = 2int_0^{2pi}int_1^3 rsqrt{r^2-1} text dr text dtheta$$



Method 3: The washer method.



Consider a horizontal washer (ring) with a thickness of $text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $text dV = (r_2^2-r_1^2)pi$. To get the total volume, integrate the volumes of all such washers:



$$V=intlimits_{-2sqrt2}^{2sqrt2} pi(9-y^2-1) text dy$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
    $endgroup$
    – Ludwig
    8 hours ago










  • $begingroup$
    @Ludwig I apologize, I have misread your question. I have updated my answer.
    $endgroup$
    – Haris Gusic
    8 hours ago
















3












$begingroup$

Your solution is correct.



Method 2: Using double integrals.



Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=sqrt{x^2+z^2-1}$, for $yge 0$. Next, define a region



$$D={(x,z) | 1le x^2+z^2 le 9}$$



To get the volume of the upper body, we evaluate the integral



$$iintlimits_D y(x,z) text dx text dz = iintlimits_D sqrt{x^2+z^2-1} text dx text dz$$



and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:



$$V = 2int_0^{2pi}int_1^3 rsqrt{r^2-1} text dr text dtheta$$



Method 3: The washer method.



Consider a horizontal washer (ring) with a thickness of $text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $text dV = (r_2^2-r_1^2)pi$. To get the total volume, integrate the volumes of all such washers:



$$V=intlimits_{-2sqrt2}^{2sqrt2} pi(9-y^2-1) text dy$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
    $endgroup$
    – Ludwig
    8 hours ago










  • $begingroup$
    @Ludwig I apologize, I have misread your question. I have updated my answer.
    $endgroup$
    – Haris Gusic
    8 hours ago














3












3








3





$begingroup$

Your solution is correct.



Method 2: Using double integrals.



Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=sqrt{x^2+z^2-1}$, for $yge 0$. Next, define a region



$$D={(x,z) | 1le x^2+z^2 le 9}$$



To get the volume of the upper body, we evaluate the integral



$$iintlimits_D y(x,z) text dx text dz = iintlimits_D sqrt{x^2+z^2-1} text dx text dz$$



and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:



$$V = 2int_0^{2pi}int_1^3 rsqrt{r^2-1} text dr text dtheta$$



Method 3: The washer method.



Consider a horizontal washer (ring) with a thickness of $text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $text dV = (r_2^2-r_1^2)pi$. To get the total volume, integrate the volumes of all such washers:



$$V=intlimits_{-2sqrt2}^{2sqrt2} pi(9-y^2-1) text dy$$






share|cite|improve this answer











$endgroup$



Your solution is correct.



Method 2: Using double integrals.



Namely, by rotating the graph around the $y$-axis, we can define $y$ as a two-variable function $y(x,z)=sqrt{x^2+z^2-1}$, for $yge 0$. Next, define a region



$$D={(x,z) | 1le x^2+z^2 le 9}$$



To get the volume of the upper body, we evaluate the integral



$$iintlimits_D y(x,z) text dx text dz = iintlimits_D sqrt{x^2+z^2-1} text dx text dz$$



and to get the total volume, we just multiply this by two. The above integral can be easily found using polar coordinates, and we have:



$$V = 2int_0^{2pi}int_1^3 rsqrt{r^2-1} text dr text dtheta$$



Method 3: The washer method.



Consider a horizontal washer (ring) with a thickness of $text dy$, at a height $y$ from the $x$-axis. Its inner radius is $r_1 = sqrt{1+y^2}$ and its outer radius is $r_2 = 3$. The volume of the washer is $text dV = (r_2^2-r_1^2)pi$. To get the total volume, integrate the volumes of all such washers:



$$V=intlimits_{-2sqrt2}^{2sqrt2} pi(9-y^2-1) text dy$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 8 hours ago

























answered 9 hours ago









Haris GusicHaris Gusic

2,553322




2,553322












  • $begingroup$
    But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
    $endgroup$
    – Ludwig
    8 hours ago










  • $begingroup$
    @Ludwig I apologize, I have misread your question. I have updated my answer.
    $endgroup$
    – Haris Gusic
    8 hours ago


















  • $begingroup$
    But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
    $endgroup$
    – Ludwig
    8 hours ago










  • $begingroup$
    @Ludwig I apologize, I have misread your question. I have updated my answer.
    $endgroup$
    – Haris Gusic
    8 hours ago
















$begingroup$
But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
$endgroup$
– Ludwig
8 hours ago




$begingroup$
But when using cylindrical shells to integrate don't you need to take the integral of $2 pi x f(x)$ where $x$ is the radius of the incremental cylinder and $f(x)$ is the height of that incremental cylinder.
$endgroup$
– Ludwig
8 hours ago












$begingroup$
@Ludwig I apologize, I have misread your question. I have updated my answer.
$endgroup$
– Haris Gusic
8 hours ago




$begingroup$
@Ludwig I apologize, I have misread your question. I have updated my answer.
$endgroup$
– Haris Gusic
8 hours ago











4












$begingroup$

The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.



$$
V=2cdot 2piint_{1}^{3}xsqrt{x^2-1},dx=
frac{4}{2}piint_{1}^{3}sqrt{x^2-1}frac{d}{dx}(x^2-1),dx=\
2piint_{0}^{8}sqrt{u},du=2pifrac{2sqrt{u^3}}{3}bigg|_{0}^{8}=
frac{4}{3}pileft(sqrt{8^3}-sqrt{0}right)=frac{64sqrt{2}pi}{3}
$$



Wolfram Alpha check






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    8 hours ago






  • 1




    $begingroup$
    I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
    $endgroup$
    – Michael Rybkin
    8 hours ago


















4












$begingroup$

The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.



$$
V=2cdot 2piint_{1}^{3}xsqrt{x^2-1},dx=
frac{4}{2}piint_{1}^{3}sqrt{x^2-1}frac{d}{dx}(x^2-1),dx=\
2piint_{0}^{8}sqrt{u},du=2pifrac{2sqrt{u^3}}{3}bigg|_{0}^{8}=
frac{4}{3}pileft(sqrt{8^3}-sqrt{0}right)=frac{64sqrt{2}pi}{3}
$$



Wolfram Alpha check






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    8 hours ago






  • 1




    $begingroup$
    I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
    $endgroup$
    – Michael Rybkin
    8 hours ago
















4












4








4





$begingroup$

The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.



$$
V=2cdot 2piint_{1}^{3}xsqrt{x^2-1},dx=
frac{4}{2}piint_{1}^{3}sqrt{x^2-1}frac{d}{dx}(x^2-1),dx=\
2piint_{0}^{8}sqrt{u},du=2pifrac{2sqrt{u^3}}{3}bigg|_{0}^{8}=
frac{4}{3}pileft(sqrt{8^3}-sqrt{0}right)=frac{64sqrt{2}pi}{3}
$$



Wolfram Alpha check






share|cite|improve this answer









$endgroup$



The way you set up the integral seems to be correct (that's the exact same way I would set it up), but I think you calculated it slightly wrong. You forgot that you also have the lower part of the hyperbola. So, the volume should be doubled.



$$
V=2cdot 2piint_{1}^{3}xsqrt{x^2-1},dx=
frac{4}{2}piint_{1}^{3}sqrt{x^2-1}frac{d}{dx}(x^2-1),dx=\
2piint_{0}^{8}sqrt{u},du=2pifrac{2sqrt{u^3}}{3}bigg|_{0}^{8}=
frac{4}{3}pileft(sqrt{8^3}-sqrt{0}right)=frac{64sqrt{2}pi}{3}
$$



Wolfram Alpha check







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









Michael RybkinMichael Rybkin

3,685420




3,685420








  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    8 hours ago






  • 1




    $begingroup$
    I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
    $endgroup$
    – Michael Rybkin
    8 hours ago
















  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    8 hours ago






  • 1




    $begingroup$
    I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
    $endgroup$
    – Michael Rybkin
    8 hours ago










1




1




$begingroup$
The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
$endgroup$
– Haris Gusic
8 hours ago




$begingroup$
The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
$endgroup$
– Haris Gusic
8 hours ago




1




1




$begingroup$
I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
$endgroup$
– Michael Rybkin
8 hours ago






$begingroup$
I think the OP lost the $1/2$ somewhere that they should have gotten from doing the u-substitution.
$endgroup$
– Michael Rybkin
8 hours ago




















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%2f3141433%2fvolume-of-hyperbola-revolved-about-the-y-axis%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







-

Popular posts from this blog

"Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness.
Read more

Alcedinidae

Image
Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
Read more

RAC Tourist Trophy

Image
Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
Read more