Volume of Solid of Revolution (Glass)
$begingroup$
We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?
Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$
integration volume solid-of-revolution
asked Dec 19 '18 at 15:22
M PapasM Papas
76
$endgroup$
|
show 5 more comments
$begingroup$
We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?
Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$
integration volume solid-of-revolution
asked Dec 19 '18 at 15:22
M PapasM Papas
76
$endgroup$
$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28
$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35
$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21
$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22
$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30
|
show 5 more comments
$begingroup$
We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?
Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$
integration volume solid-of-revolution
asked Dec 19 '18 at 15:22
M PapasM Papas
76
$endgroup$
We are given $y^2/a^2-x^2/b^2=1$, $y>0$ . If we rotate the hyperbola around the $y$ axis the shape is similar to a glass.
What will the volume of water inside the glass be, in order to fill the glass up to $y=A$, $A>a$?
Following the given hint I found
$$x=b*(y^2/a^2-1)^{1/2}.$$
integration volume solid-of-revolution
integration volume solid-of-revolution
asked Dec 19 '18 at 15:22
M PapasM Papas
76
asked Dec 19 '18 at 15:22
M PapasM Papas
76
edited Dec 20 '18 at 13:06
M Papas
asked Dec 19 '18 at 15:22
M PapasM Papas
76
asked Dec 19 '18 at 15:22
M PapasM Papas
76
asked Dec 19 '18 at 15:22
M PapasM Papas
76
76
$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28
$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35
$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21
$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22
$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30
|
show 5 more comments
$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28
$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35
$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21
$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22
$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30
$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28
$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
$endgroup$
– Robert Z
Dec 19 '18 at 15:28
$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35
$begingroup$
See my hint below. Now you should know how to start.
$endgroup$
– Robert Z
Dec 19 '18 at 15:35
$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21
$begingroup$
Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
$endgroup$
– Andrei
Dec 19 '18 at 16:21
$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22
$begingroup$
I have done this using geogebra, however I am not sure what my integral should be
$endgroup$
– M Papas
Dec 19 '18 at 16:22
$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30
$begingroup$
Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
$endgroup$
– Andrei
Dec 19 '18 at 16:30
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
$endgroup$
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
add a comment |
$begingroup$
I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
$endgroup$
add a comment |
$begingroup$
So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
$endgroup$
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
add a comment |
$begingroup$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
$endgroup$
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
add a comment |
$begingroup$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
$endgroup$
The green axis in your picture is the $y$ axis. You can calculate the volume in two ways: disks or shells.
The disks are perpendicular to the $y$ axis, have a thickness $dy$ and the radius is equal to $x$ at a given $y$. In this case your integration variable is $y$. So the volume of the disk is $pi x^2 dy=b^2(y^2/a^2-1) dy$. You know the minimum $y$ is $a$, and the maximum is $A$.
The shell method means that you need to have cylindrical shells, with the axis parallel to the $y$ axis. The radius of the cylinder is $x$, the thickness is $dx$. The volume of the shell is then $2pi x h(x)mathrm dx$, where $h(x)$ is the height of the shell. The "bottom" of the shell starts at $y=asqrt{x^2/b^2+1}$, and the "top" is at $y=A$, so $h(x)=A-asqrt{x^2/b^2+1}$ The integration range starts from $0$ and goes on until the height is $0$, or $y(x)=A$.
$$frac{A^2}{a^2}-frac{x_{max}^2}{b^2}=1$$
You can now get the upper limit of $x$ from the above equation.
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
answered Dec 19 '18 at 16:50
AndreiAndrei
13.3k21230
13.3k21230
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
add a comment |
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
which integral should I use to calculate the volume?
$endgroup$
– M Papas
Dec 19 '18 at 16:53
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
Do both. Just to check that you get the same answer. Personally, I think the disk method is easier in this case. Integrating polynomials is easier that integrating functions with square roots.
$endgroup$
– Andrei
Dec 19 '18 at 16:57
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
In the disk method what should my integral be? I didint really get it
$endgroup$
– M Papas
Dec 19 '18 at 17:02
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
$begingroup$
You integrate over $y$, between $a$ and $A$ the volume of the disk
$endgroup$
– Andrei
Dec 19 '18 at 17:07
add a comment |
$begingroup$
I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
$endgroup$
add a comment |
$begingroup$
I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
$endgroup$
add a comment |
$begingroup$
I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
$endgroup$
I do the problem in cylindrical coordinates as follows:
Converting the equation of hyperbola to hyperboloid around $z$ can be simply done by replacing $yto z$ and $xto r$ as: $$frac{z^2}{a^2}-frac{r^2}{b^2}=1$$
(I figured this by reconciling $r$ with $x$ in the $x-z$ plane)
Now the integral becomes terribly simple:
$$int_0^{2pi}int_a^Aint_0^{sqrt{b^2(frac{z^2}{a^2}-1)}} rdrdzdtheta=pi b^2int_a^ABigg(frac{z^2}{a^2}-1Bigg)dz=pi b^2Bigg(frac{(A^3-a^3)}{3a^2}-(A-a)Bigg)$$
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
edited Dec 19 '18 at 17:17
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
answered Dec 19 '18 at 17:11
Sameer BahetiSameer Baheti
5618
5618
add a comment |
add a comment |
$begingroup$
So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
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So in your drawing you show that you are almost near to find the good start.
Take half of the trapezoidal stripe you indicated, going from the y axis to the hyperbola.
Its width is $dy$, one base is $x(y)$ and the other $x(y+dy)=x(y)+x'(y)dy$.
When you rotate it by $2pi$ you get a conic frustum of height $dy$ and volume $pi x(y)^2 dy$ because you know that higher order infinitesimal can be waived.
Then just integrate the above along $y$, between the bounds you need.
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
answered Dec 19 '18 at 16:55
G CabG Cab
20.4k31341
20.4k31341
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$begingroup$
Why don't you show your attempt? Otherwise we are not able to help you.
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– Robert Z
Dec 19 '18 at 15:28
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See my hint below. Now you should know how to start.
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– Robert Z
Dec 19 '18 at 15:35
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Start by drawing a picture of the hyperbola. See where you have the intersection with $y=A$.
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– Andrei
Dec 19 '18 at 16:21
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I have done this using geogebra, however I am not sure what my integral should be
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– M Papas
Dec 19 '18 at 16:22
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Why don't you paste the picture (or a link to it) into your question? It would help us to reference it in our answers.
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– Andrei
Dec 19 '18 at 16:30