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













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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
    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




















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Can an atomic nucleus contain both particles and antiparticles? [duplicate]

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9 $begingroup$ This question already has an answer here: What happens if we put together a proton and an antineutron? 2 answers Is it theoretically possible to make a "deuterium" atom containing a proton and an antineutron in its nucleus? Would the strong nuclear force cause attraction between a proton and an antineutron? Would such a nucleus be stable, or would the proton somehow annihilate the antineutron when close enough? nuclear-physics antimatter share | cite | improve this question asked yesterday
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