Apostol's Calculus volume 1, 2.13 exercise 16
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This is a problem from Apostol's calculus, I have figured out how to find the volume, but not sure about the sketch, what am I supposed to sketch exactly? The problem has not given me what the solid is?
"The cross sections of a solid are squares perpendicular to the x-axis with their centers on the axis. If the square cut off at x has edge $2x^{2}$, find the volume of the solid between $x = 0$ and $x = a$. Make a sketch."
calculus
asked Jan 5 at 17:36
D. QaD. Qa
1656
$endgroup$
add a comment |
$begingroup$
This is a problem from Apostol's calculus, I have figured out how to find the volume, but not sure about the sketch, what am I supposed to sketch exactly? The problem has not given me what the solid is?
"The cross sections of a solid are squares perpendicular to the x-axis with their centers on the axis. If the square cut off at x has edge $2x^{2}$, find the volume of the solid between $x = 0$ and $x = a$. Make a sketch."
calculus
asked Jan 5 at 17:36
D. QaD. Qa
1656
$endgroup$
$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56
add a comment |
$begingroup$
This is a problem from Apostol's calculus, I have figured out how to find the volume, but not sure about the sketch, what am I supposed to sketch exactly? The problem has not given me what the solid is?
"The cross sections of a solid are squares perpendicular to the x-axis with their centers on the axis. If the square cut off at x has edge $2x^{2}$, find the volume of the solid between $x = 0$ and $x = a$. Make a sketch."
calculus
asked Jan 5 at 17:36
D. QaD. Qa
1656
$endgroup$
This is a problem from Apostol's calculus, I have figured out how to find the volume, but not sure about the sketch, what am I supposed to sketch exactly? The problem has not given me what the solid is?
"The cross sections of a solid are squares perpendicular to the x-axis with their centers on the axis. If the square cut off at x has edge $2x^{2}$, find the volume of the solid between $x = 0$ and $x = a$. Make a sketch."
calculus
calculus
asked Jan 5 at 17:36
D. QaD. Qa
1656
asked Jan 5 at 17:36
D. QaD. Qa
1656
asked Jan 5 at 17:36
D. QaD. Qa
1656
asked Jan 5 at 17:36
D. QaD. Qa
1656
asked Jan 5 at 17:36
D. QaD. Qa
1656
1656
$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56
add a comment |
$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56
$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The volume is bounded by the following curves:
$z = x^2, z = -x^2, y = x^2, y = -x^2$ and x from 0 to a.
There are two parabolas one on positive z axis and one on the negative z axis. Similarly one on the positive y axis and one on the negative y axis. Hence when you view along the x axis towards the origin, you find a tunnel with square cross sections at each x value.
answered Jan 23 at 1:32
KY TangKY Tang
50436
$endgroup$
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The volume is bounded by the following curves:
$z = x^2, z = -x^2, y = x^2, y = -x^2$ and x from 0 to a.
There are two parabolas one on positive z axis and one on the negative z axis. Similarly one on the positive y axis and one on the negative y axis. Hence when you view along the x axis towards the origin, you find a tunnel with square cross sections at each x value.
answered Jan 23 at 1:32
KY TangKY Tang
50436
$endgroup$
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
add a comment |
$begingroup$
The volume is bounded by the following curves:
$z = x^2, z = -x^2, y = x^2, y = -x^2$ and x from 0 to a.
There are two parabolas one on positive z axis and one on the negative z axis. Similarly one on the positive y axis and one on the negative y axis. Hence when you view along the x axis towards the origin, you find a tunnel with square cross sections at each x value.
answered Jan 23 at 1:32
KY TangKY Tang
50436
$endgroup$
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
add a comment |
$begingroup$
The volume is bounded by the following curves:
$z = x^2, z = -x^2, y = x^2, y = -x^2$ and x from 0 to a.
There are two parabolas one on positive z axis and one on the negative z axis. Similarly one on the positive y axis and one on the negative y axis. Hence when you view along the x axis towards the origin, you find a tunnel with square cross sections at each x value.
answered Jan 23 at 1:32
KY TangKY Tang
50436
$endgroup$
The volume is bounded by the following curves:
$z = x^2, z = -x^2, y = x^2, y = -x^2$ and x from 0 to a.
There are two parabolas one on positive z axis and one on the negative z axis. Similarly one on the positive y axis and one on the negative y axis. Hence when you view along the x axis towards the origin, you find a tunnel with square cross sections at each x value.
answered Jan 23 at 1:32
KY TangKY Tang
50436
edited Jan 23 at 3:10
answered Jan 23 at 1:32
KY TangKY Tang
50436
answered Jan 23 at 1:32
KY TangKY Tang
50436
answered Jan 23 at 1:32
KY TangKY Tang
50436
50436
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
add a comment |
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
$begingroup$
Sorry for the late reply, I only now noticed your answer. I tried sketching this using GeoGebra but when viewing along the x-axis towards the origin I didn't see a tunnel with square cross sections.. I am very confused
$endgroup$
– D. Qa
Jan 27 at 12:01
1
1
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
In fact the "Tunnel" is closed at the origin. When you the tunnel at a certain value of x, say 5, the first equation will give you z = 25 which is a straight line segment. Similarly, the second equation will give you z = -25 which is another straight line segment but this time in the lower quadrant. The third equation will give you y = 25 which is another line segment vertically. The fourth equation will give you another line segment y = -25. This 4 line segments will give you a square of line 50 or 2x^2. This is true for any value of x when x > 0.
$endgroup$
– KY Tang
Jan 27 at 18:40
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
$begingroup$
Although I am still having trouble to graph it, but I think I somehow understood what you meant.
$endgroup$
– D. Qa
Jan 27 at 18:54
add a comment |
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$begingroup$
and by similar triangles I'd get $s= dfrac{L}{h}cdot x= dfrac{2x^{2}}{x} cdot x = 2x^{2}$ ? I don't get how did we know from this problem that the length of the base would be $2x^{2}$?
$endgroup$
– D. Qa
Jan 7 at 16:20
$begingroup$
I retract my original interpretation. The figure must have curved sides since the cross-section $x$ units to the right of the origin has vertices $(x, x^2, x^2)$, $(x, x^2, -x^2)$, $(x, -x^2, -x^2)$, and $(x, -x^2, x^2)$.
$endgroup$
– N. F. Taussig
Jan 8 at 10:29
$begingroup$
I am sorry, but I am kinda confused. If we were to draw it 2D would it be a parabola? then we'd imagine it has a square cross section?
$endgroup$
– D. Qa
Jan 8 at 20:27
$begingroup$
If you were to take a cross section along the $x$-axis, the figure would lie between two parabolas that bend away from the axis.
$endgroup$
– N. F. Taussig
Jan 8 at 20:56