Triple integral - wedge shaped solid
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Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
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up vote
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down vote
favorite
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
118112
add a comment |Â
add a comment |Â
3 Answers
3
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oldest
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0
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looks like this should be the required vol
$$2int_0^pi/2 int_0^2 int_0^rcostheta rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,590821
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
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I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
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Planes $x=rm const.$ intersect this solid $B$ in rectangles of width $2sqrt4-x^2$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$$rm vol(B)=int_0^2 2sqrt4-x^2>x>dx=-2over3(4-x^2)^3/2biggr|_0^2=16over3 .$$
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
looks like this should be the required vol
$$2int_0^pi/2 int_0^2 int_0^rcostheta rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,590821
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
add a comment |Â
up vote
0
down vote
looks like this should be the required vol
$$2int_0^pi/2 int_0^2 int_0^rcostheta rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,590821
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
looks like this should be the required vol
$$2int_0^pi/2 int_0^2 int_0^rcostheta rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,590821
looks like this should be the required vol
$$2int_0^pi/2 int_0^2 int_0^rcostheta rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,590821
answered Mar 28 '14 at 13:06
ketan
1,590821
answered Mar 28 '14 at 13:06
ketan
1,590821
answered Mar 28 '14 at 13:06
ketan
1,590821
1,590821
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
add a comment |Â
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
why is it [$0,fracpi2$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,fracpi2]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_-pi/2^pi/2$.
– orion
Mar 28 '14 at 13:18
add a comment |Â
up vote
0
down vote
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
add a comment |Â
up vote
0
down vote
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
answered Jun 7 '17 at 17:04
Cye Waldman
4,0252522
4,0252522
add a comment |Â
add a comment |Â
up vote
0
down vote
Planes $x=rm const.$ intersect this solid $B$ in rectangles of width $2sqrt4-x^2$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$$rm vol(B)=int_0^2 2sqrt4-x^2>x>dx=-2over3(4-x^2)^3/2biggr|_0^2=16over3 .$$
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
add a comment |Â
up vote
0
down vote
Planes $x=rm const.$ intersect this solid $B$ in rectangles of width $2sqrt4-x^2$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$$rm vol(B)=int_0^2 2sqrt4-x^2>x>dx=-2over3(4-x^2)^3/2biggr|_0^2=16over3 .$$
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Planes $x=rm const.$ intersect this solid $B$ in rectangles of width $2sqrt4-x^2$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$$rm vol(B)=int_0^2 2sqrt4-x^2>x>dx=-2over3(4-x^2)^3/2biggr|_0^2=16over3 .$$
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
Planes $x=rm const.$ intersect this solid $B$ in rectangles of width $2sqrt4-x^2$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$$rm vol(B)=int_0^2 2sqrt4-x^2>x>dx=-2over3(4-x^2)^3/2biggr|_0^2=16over3 .$$
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
answered Jul 9 '17 at 9:08
Christian Blatter
165k7109311
165k7109311
add a comment |Â
add a comment |Â
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