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Surface area of revolution of trig fn
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I want to fine the surface area of revoultion around the y-axis of
$$x=cos^3(theta), y=sin^3(theta)$$
I looked up a answer below but it's confusing me. (note that it's revolution around x-axis)
To find the surface area, shouldn't it be $ S=int_a^b 2pi y ;dx $ ? similar to disk method?
where does the $sqrt(dx)^2+(dy)^2$ come from? I believe it has to do with line integral and have no idea what it does here.
definite-integrals
asked Jul 26 at 6:26
NK Yu
1847
add a comment |Â
up vote
0
down vote
favorite
I want to fine the surface area of revoultion around the y-axis of
$$x=cos^3(theta), y=sin^3(theta)$$
I looked up a answer below but it's confusing me. (note that it's revolution around x-axis)
To find the surface area, shouldn't it be $ S=int_a^b 2pi y ;dx $ ? similar to disk method?
where does the $sqrt(dx)^2+(dy)^2$ come from? I believe it has to do with line integral and have no idea what it does here.
definite-integrals
asked Jul 26 at 6:26
NK Yu
1847
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to fine the surface area of revoultion around the y-axis of
$$x=cos^3(theta), y=sin^3(theta)$$
I looked up a answer below but it's confusing me. (note that it's revolution around x-axis)
To find the surface area, shouldn't it be $ S=int_a^b 2pi y ;dx $ ? similar to disk method?
where does the $sqrt(dx)^2+(dy)^2$ come from? I believe it has to do with line integral and have no idea what it does here.
definite-integrals
asked Jul 26 at 6:26
NK Yu
1847
I want to fine the surface area of revoultion around the y-axis of
$$x=cos^3(theta), y=sin^3(theta)$$
I looked up a answer below but it's confusing me. (note that it's revolution around x-axis)
To find the surface area, shouldn't it be $ S=int_a^b 2pi y ;dx $ ? similar to disk method?
where does the $sqrt(dx)^2+(dy)^2$ come from? I believe it has to do with line integral and have no idea what it does here.
definite-integrals
asked Jul 26 at 6:26
NK Yu
1847
asked Jul 26 at 6:26
NK Yu
1847
asked Jul 26 at 6:26
NK Yu
1847
asked Jul 26 at 6:26
NK Yu
1847
1847
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is
$$A=pi(r_1+r_2)ell ,$$
where $r_1$ and $r_2$ are the radii of the two ends and $ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is
$$dell=sqrt(dx)^2+(dy)^2
=sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2
dtheta .$$
So the area element is
$$dA=2pi f(x)sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2dtheta .$$
answered Jul 26 at 6:45
David
65.8k662124
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is
$$A=pi(r_1+r_2)ell ,$$
where $r_1$ and $r_2$ are the radii of the two ends and $ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is
$$dell=sqrt(dx)^2+(dy)^2
=sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2
dtheta .$$
So the area element is
$$dA=2pi f(x)sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2dtheta .$$
answered Jul 26 at 6:45
David
65.8k662124
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
add a comment |Â
up vote
1
down vote
accepted
If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is
$$A=pi(r_1+r_2)ell ,$$
where $r_1$ and $r_2$ are the radii of the two ends and $ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is
$$dell=sqrt(dx)^2+(dy)^2
=sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2
dtheta .$$
So the area element is
$$dA=2pi f(x)sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2dtheta .$$
answered Jul 26 at 6:45
David
65.8k662124
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is
$$A=pi(r_1+r_2)ell ,$$
where $r_1$ and $r_2$ are the radii of the two ends and $ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is
$$dell=sqrt(dx)^2+(dy)^2
=sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2
dtheta .$$
So the area element is
$$dA=2pi f(x)sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2dtheta .$$
answered Jul 26 at 6:45
David
65.8k662124
If you take a slice of the surface of revolution perpendicular to its axis, the shape you get will be (in the limit of an infinitely thin slice) part of a cone, not of a cylinder. The area of such a shape is
$$A=pi(r_1+r_2)ell ,$$
where $r_1$ and $r_2$ are the radii of the two ends and $ell$ is the slant height (not the vertical or horizontal "height"). In the limit, $r_1$ and $r_2$ are both $f(x)$, and the slant height is
$$dell=sqrt(dx)^2+(dy)^2
=sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2
dtheta .$$
So the area element is
$$dA=2pi f(x)sqrtBigl(fracdxdthetaBigr)^2+Bigl(fracdydthetaBigr)^2dtheta .$$
answered Jul 26 at 6:45
David
65.8k662124
answered Jul 26 at 6:45
David
65.8k662124
answered Jul 26 at 6:45
David
65.8k662124
answered Jul 26 at 6:45
David
65.8k662124
65.8k662124
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
add a comment |Â
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
I think I got it. I'll try it myself. Thank you!
– NK Yu
Jul 26 at 7:30
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
How come r1+r2 = 2f(x) instead of f(x)+f(x+dx) or y+y+dy?
– NK Yu
Jul 26 at 7:42
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
In the limit as the slice becomes infinitely thin, there is no difference between $y$ and $y+dy$.
– David
Jul 26 at 23:17
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
These kind of concepts always confuse me. they are sometimes the same yet sometimes different a little bit. am I going too far if I ask you why they are said to be the same?
– NK Yu
Jul 27 at 12:06
add a comment |Â
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