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
    8 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    8 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
    8 hours ago










  • $begingroup$
    Yes, sorry, the bounds on the x values are 1 and 3
    $endgroup$
    – Ludwig
    8 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 8 hours ago







Ludwig

















asked 8 hours ago









LudwigLudwig

805




805












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










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


















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










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
















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




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












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




$begingroup$
Yes, sorry, the bounds on the x values are 1 and 3
$endgroup$
– Ludwig
8 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
    7 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
    7 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
    7 hours ago













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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
    7 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
    7 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 7 hours ago

























answered 8 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
    7 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
    7 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
7 hours ago




$begingroup$
@Ludwig I apologize, I have misread your question. I have updated my answer.
$endgroup$
– Haris Gusic
7 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
    7 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
    7 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
    7 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
    7 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 7 hours ago









Michael RybkinMichael Rybkin

3,685420




3,685420








  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    7 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
    7 hours ago
















  • 1




    $begingroup$
    The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
    $endgroup$
    – Haris Gusic
    7 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
    7 hours ago










1




1




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




$begingroup$
The OP's solution is equivalent, since $sqrt8 =2sqrt2$.
$endgroup$
– Haris Gusic
7 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
7 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
7 hours ago




















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CARDNET

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CARDNET は日本カードネットワークが運営するクレジット決済ネットワーク。日本カードネットワークの略称としてもCARDNETが使用されている。 事件・事故 2017年4月15日午前11時8分からクレジットカードの決済がしづらくなる障害が発生した。障害は同日午後5時18分に復旧した。原因は、複数あるCARDNETセンターの拠点のうち1つで、2重化してあるL3スイッチの片方が故障し、1系統にトラフィックが集中したことによる輻輳が発生したため。対処としては、故障したL3スイッチ交換とシステム再起動が行われた [1] 。 出典 ^ 金子寛人 ( 2017年4月17日 ). “CARDNETのクレジット決済に6時間強障害、原因はL3スイッチ故障”. 日経コンピュータ. 2018年10月18日 閲覧。 この項目は、コンピュータに関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（PJ:コンピュータ/P:コンピュータ）。 この項目は、経済に関連した 書きかけの項目 です。この項目を加筆・訂正などしてくださる協力者を求めています（ポータル 経済学、プロジェクト 経済）。 This page is only for reference, If you need detailed information, please check here
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Boot-repair Failure: Unable to locate package grub-common:i386

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0 I am on a system dual-booting Windows and Debian (usually using the latter), and a recent Windows update appears to have wreaked havoc on the Debian boot system. I managed to boot Windows again, but not Linux. Following online suggestions, I created a Live USB of Ubuntu, and booted from that. Then I have installed and run the boot-repair utility. The utility instructs me to type several lines including sudo chroot "/mnt/boot-sav/sda6" apt-get purge -y grub*-common grub-common:i386 To which I receive the error E: Unable to locate package grub-common:i386 I am given to understand the issue may have to do with having UEFI vs BIOS (I believe I have UEFI), but do not understand this well. In any case I have subsequently run the boot-repair BootInfo Summary utility which created the following file h
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濃尾地震

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濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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