Mixing
Volumes of Solids of Revolution - I used two methods and got two different answers, could someone check where...
$begingroup$
The question is as follows:
$f(x) = x^2$ and $g(x) = -x^2$, rotated about the y-axis. It is also bounded by $x = -2$ Find the volume by rotating the region around the given axis.
The first method I tried was "shell" method. My equation was:
$$2piint_{-2}^0-x(2x^2)dx $$
The second method (according to my teacher) was to take the region in Quadrant II, rotate it and find volume, then multiply by 2 (using disk method):
$$ 2timespiint_{-2}^0 x^4dx$$
And of course they ended up being different results. I'm not sure if I messed up in shell method equation, second method equation, or both, so I would appreciate it if someone offered correction.
calculus definite-integrals
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
$endgroup$
add a comment |
$begingroup$
The question is as follows:
$f(x) = x^2$ and $g(x) = -x^2$, rotated about the y-axis. It is also bounded by $x = -2$ Find the volume by rotating the region around the given axis.
The first method I tried was "shell" method. My equation was:
$$2piint_{-2}^0-x(2x^2)dx $$
The second method (according to my teacher) was to take the region in Quadrant II, rotate it and find volume, then multiply by 2 (using disk method):
$$ 2timespiint_{-2}^0 x^4dx$$
And of course they ended up being different results. I'm not sure if I messed up in shell method equation, second method equation, or both, so I would appreciate it if someone offered correction.
calculus definite-integrals
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
$endgroup$
$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
2
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
1
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50
add a comment |
$begingroup$
The question is as follows:
$f(x) = x^2$ and $g(x) = -x^2$, rotated about the y-axis. It is also bounded by $x = -2$ Find the volume by rotating the region around the given axis.
The first method I tried was "shell" method. My equation was:
$$2piint_{-2}^0-x(2x^2)dx $$
The second method (according to my teacher) was to take the region in Quadrant II, rotate it and find volume, then multiply by 2 (using disk method):
$$ 2timespiint_{-2}^0 x^4dx$$
And of course they ended up being different results. I'm not sure if I messed up in shell method equation, second method equation, or both, so I would appreciate it if someone offered correction.
calculus definite-integrals
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
$endgroup$
The question is as follows:
$f(x) = x^2$ and $g(x) = -x^2$, rotated about the y-axis. It is also bounded by $x = -2$ Find the volume by rotating the region around the given axis.
The first method I tried was "shell" method. My equation was:
$$2piint_{-2}^0-x(2x^2)dx $$
The second method (according to my teacher) was to take the region in Quadrant II, rotate it and find volume, then multiply by 2 (using disk method):
$$ 2timespiint_{-2}^0 x^4dx$$
And of course they ended up being different results. I'm not sure if I messed up in shell method equation, second method equation, or both, so I would appreciate it if someone offered correction.
calculus definite-integrals
calculus definite-integrals
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
edited Jan 28 at 23:43
Andrew Wang
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
asked Jan 28 at 23:35
Andrew WangAndrew Wang
415
415
$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
2
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
1
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50
add a comment |
$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
2
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
1
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50
$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
2
2
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
1
1
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50
add a comment |
1 Answer
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$begingroup$
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=int_0^24pi x^3,dx=16pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=sqrt{y}$. So we get a volume of
$$ V=int_{-4}^4pi(2^2-(sqrt{y})^2),dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2piint_{0}^4(4-y),dy=16pi $$
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=int_0^24pi x^3,dx=16pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=sqrt{y}$. So we get a volume of
$$ V=int_{-4}^4pi(2^2-(sqrt{y})^2),dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2piint_{0}^4(4-y),dy=16pi $$
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=int_0^24pi x^3,dx=16pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=sqrt{y}$. So we get a volume of
$$ V=int_{-4}^4pi(2^2-(sqrt{y})^2),dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2piint_{0}^4(4-y),dy=16pi $$
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
add a comment |
$begingroup$
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=int_0^24pi x^3,dx=16pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=sqrt{y}$. So we get a volume of
$$ V=int_{-4}^4pi(2^2-(sqrt{y})^2),dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2piint_{0}^4(4-y),dy=16pi $$
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
The following diagram is a representation of the solid of revolution.
Because of the symmetry, you can integrate along the horizontal axis. from $0$ to $2$ rather than from $-2$ to $0$.
For the cylindrical method, the cylindrical sections have area
$$ A=2pi rh $$
In the cylindrical shell method the cross-sections through the region are taken parallel to the axis of revolution. Thus the cross-sections are perpendicular to the $x$-axis. So we will integrate with respect to $x$ and find $r$ and $h$ as functions of $x$.
For this example, $r=x$ and $h=2x^2$ so the volume is
$$ V=int_0^24pi x^3,dx=16pi $$
For the annulus method, we take a cross-section of the region perpendicular to the axis of revolution. These cross-sections are perpendicular to the $y$-axis so we will integrate with respect to $y$ and find the variables of the area formula
$$A=pi(R^2-r^2)$$
in terms of $y$. We see that the outer radius is $R=2$ and the inner radius is $r=sqrt{y}$. So we get a volume of
$$ V=int_{-4}^4pi(2^2-(sqrt{y})^2),dy $$
but we can use your teacher's suggestion and just double the volume of the top half:
$$ V=2piint_{0}^4(4-y),dy=16pi $$
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
edited Jan 30 at 3:36
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
answered Jan 30 at 3:17
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
add a comment |
add a comment |
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$begingroup$
Where do $-2$ and $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:37
$begingroup$
@JoséCarlosSantos Oh sorry I forgot to add the requirement x = -2 in the text. I will update the question accordingly.
$endgroup$
– Andrew Wang
Jan 28 at 23:42
2
$begingroup$
Still, where does the $0$ come from?
$endgroup$
– José Carlos Santos
Jan 28 at 23:43
1
$begingroup$
The shell method is used correctly, but for the disk method, which in this case could be called the annulus method, the integral should be $2timesint_0^4 A_y, dy$, where $A_y$ is the area of the intersection of the solid of revolution and the plane at height $y$.
$endgroup$
– random
Jan 29 at 0:50