I don't understand a part of the solution using the method of shells on this problem
$begingroup$
begin{array} { l } { 4 mathrm { C } - 3 text { Find the volume of the region } sqrt { x } leq y leq 1 , x geq 0 text { revolved around the } y text { -axis } } \ { text { by both the method of shells and the method of disks and washers. } } end{array}
That is the problem. It is from MIT's OCW. It has a solution, which is
begin{array} { l } { 4 mathrm { C } - 3 text { Shells: } quad int _ { 0 } ^ { 1 } 2 pi x ( 1 - y ) d x = int _ { 0 } ^ { 1 } 2 pi x ( 1 - sqrt { x } ) d x = pi / 5} end{array}
I don't understand why it's (1−y) and not just y.
calculus integration solid-of-revolution
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
$endgroup$
add a comment |
$begingroup$
begin{array} { l } { 4 mathrm { C } - 3 text { Find the volume of the region } sqrt { x } leq y leq 1 , x geq 0 text { revolved around the } y text { -axis } } \ { text { by both the method of shells and the method of disks and washers. } } end{array}
That is the problem. It is from MIT's OCW. It has a solution, which is
begin{array} { l } { 4 mathrm { C } - 3 text { Shells: } quad int _ { 0 } ^ { 1 } 2 pi x ( 1 - y ) d x = int _ { 0 } ^ { 1 } 2 pi x ( 1 - sqrt { x } ) d x = pi / 5} end{array}
I don't understand why it's (1−y) and not just y.
calculus integration solid-of-revolution
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
$endgroup$
add a comment |
$begingroup$
begin{array} { l } { 4 mathrm { C } - 3 text { Find the volume of the region } sqrt { x } leq y leq 1 , x geq 0 text { revolved around the } y text { -axis } } \ { text { by both the method of shells and the method of disks and washers. } } end{array}
That is the problem. It is from MIT's OCW. It has a solution, which is
begin{array} { l } { 4 mathrm { C } - 3 text { Shells: } quad int _ { 0 } ^ { 1 } 2 pi x ( 1 - y ) d x = int _ { 0 } ^ { 1 } 2 pi x ( 1 - sqrt { x } ) d x = pi / 5} end{array}
I don't understand why it's (1−y) and not just y.
calculus integration solid-of-revolution
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
$endgroup$
begin{array} { l } { 4 mathrm { C } - 3 text { Find the volume of the region } sqrt { x } leq y leq 1 , x geq 0 text { revolved around the } y text { -axis } } \ { text { by both the method of shells and the method of disks and washers. } } end{array}
That is the problem. It is from MIT's OCW. It has a solution, which is
begin{array} { l } { 4 mathrm { C } - 3 text { Shells: } quad int _ { 0 } ^ { 1 } 2 pi x ( 1 - y ) d x = int _ { 0 } ^ { 1 } 2 pi x ( 1 - sqrt { x } ) d x = pi / 5} end{array}
I don't understand why it's (1−y) and not just y.
calculus integration solid-of-revolution
calculus integration solid-of-revolution
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Jan 7 at 20:12
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Jan 7 at 20:09
T jeyT jey
102
asked Jan 7 at 20:09
T jeyT jey
102
asked Jan 7 at 20:09
T jeyT jey
102
102
add a comment |
add a comment |
2 Answers
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oldest
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$begingroup$
The region integrated, in the y-direction, is between $sqrt{x}$ and $1$. If the region being revolved was $0leq yleqsqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
$endgroup$
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
add a comment |
$begingroup$
If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $sqrt r$, which would be below the first surface, but $1-sqrt r$ which is between the two surfaces.
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
The region integrated, in the y-direction, is between $sqrt{x}$ and $1$. If the region being revolved was $0leq yleqsqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
$endgroup$
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
add a comment |
$begingroup$
The region integrated, in the y-direction, is between $sqrt{x}$ and $1$. If the region being revolved was $0leq yleqsqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
$endgroup$
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
add a comment |
$begingroup$
The region integrated, in the y-direction, is between $sqrt{x}$ and $1$. If the region being revolved was $0leq yleqsqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
$endgroup$
The region integrated, in the y-direction, is between $sqrt{x}$ and $1$. If the region being revolved was $0leq yleqsqrt{x}$, then it would be $y$ in the integrand. But when you're integrating two curves, it's top - bottom, which in this case is $1-y$.
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
answered Jan 7 at 20:18
Calvin GodfreyCalvin Godfrey
568311
568311
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
add a comment |
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
$begingroup$
Thank you, that explains it. :)
$endgroup$
– T jey
Jan 7 at 20:30
add a comment |
$begingroup$
If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $sqrt r$, which would be below the first surface, but $1-sqrt r$ which is between the two surfaces.
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
$endgroup$
add a comment |
$begingroup$
If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $sqrt r$, which would be below the first surface, but $1-sqrt r$ which is between the two surfaces.
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
$endgroup$
add a comment |
$begingroup$
If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $sqrt r$, which would be below the first surface, but $1-sqrt r$ which is between the two surfaces.
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
$endgroup$
If you would draw a picture, in cylindrical coordinates, the volume that you want to integrate is above the surface $h=sqrt r$, but below the surface $h=1$. So the height of the cylindrical shell is not $sqrt r$, which would be below the first surface, but $1-sqrt r$ which is between the two surfaces.
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
answered Jan 7 at 20:18
AndreiAndrei
11.6k21026
11.6k21026
add a comment |
add a comment |
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