Mixing
Volume of solid of revolution about x-axis via Disks
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Good morning, I am attempting to solve the following problem from the back of my book, but am struggling with the correct approach to setup as I cannot find Youtube videos with examples of how to work with curves involving absolute values.
The volume of the region bounded by $y = |x − 3|$ and $y = 3$ revolved about the
$x$-axis.
What I have so far:
Attempting to approach this via disk method: $pi$ * [integral from b to a of $y_1^2 - y_2^2]$
The absolute value however throws me off. Does this mean I need to consider both $x+3$ and $x-3$?
Any guidance on how to setup this integral would be greatly appreciated.
Regards,
J
calculus volume absolute-value
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
asked Jul 18 at 17:41
jackbenimbo
677
add a comment |Â
up vote
1
down vote
favorite
Good morning, I am attempting to solve the following problem from the back of my book, but am struggling with the correct approach to setup as I cannot find Youtube videos with examples of how to work with curves involving absolute values.
The volume of the region bounded by $y = |x − 3|$ and $y = 3$ revolved about the
$x$-axis.
What I have so far:
Attempting to approach this via disk method: $pi$ * [integral from b to a of $y_1^2 - y_2^2]$
The absolute value however throws me off. Does this mean I need to consider both $x+3$ and $x-3$?
Any guidance on how to setup this integral would be greatly appreciated.
Regards,
J
calculus volume absolute-value
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
asked Jul 18 at 17:41
jackbenimbo
677
1
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Good morning, I am attempting to solve the following problem from the back of my book, but am struggling with the correct approach to setup as I cannot find Youtube videos with examples of how to work with curves involving absolute values.
The volume of the region bounded by $y = |x − 3|$ and $y = 3$ revolved about the
$x$-axis.
What I have so far:
Attempting to approach this via disk method: $pi$ * [integral from b to a of $y_1^2 - y_2^2]$
The absolute value however throws me off. Does this mean I need to consider both $x+3$ and $x-3$?
Any guidance on how to setup this integral would be greatly appreciated.
Regards,
J
calculus volume absolute-value
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
asked Jul 18 at 17:41
jackbenimbo
677
Good morning, I am attempting to solve the following problem from the back of my book, but am struggling with the correct approach to setup as I cannot find Youtube videos with examples of how to work with curves involving absolute values.
The volume of the region bounded by $y = |x − 3|$ and $y = 3$ revolved about the
$x$-axis.
What I have so far:
Attempting to approach this via disk method: $pi$ * [integral from b to a of $y_1^2 - y_2^2]$
The absolute value however throws me off. Does this mean I need to consider both $x+3$ and $x-3$?
Any guidance on how to setup this integral would be greatly appreciated.
Regards,
J
calculus volume absolute-value
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
asked Jul 18 at 17:41
jackbenimbo
677
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
edited Jul 18 at 17:55
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 18 at 17:41
jackbenimbo
677
asked Jul 18 at 17:41
jackbenimbo
677
asked Jul 18 at 17:41
jackbenimbo
677
677
1
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06
add a comment |Â
1
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06
1
1
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The region revolved looks like this:
^y
|
3#######
|#####
| ###
o--#--->x
3
Consider each side of the absolute value separately. For this region's left half with $0le xle3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3le xle6$, the lower bound changes to $y=x-3$.
The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation.
$$piint_0^3(3^2-(3-x)^2),dx$$
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
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
The region revolved looks like this:
^y
|
3#######
|#####
| ###
o--#--->x
3
Consider each side of the absolute value separately. For this region's left half with $0le xle3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3le xle6$, the lower bound changes to $y=x-3$.
The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation.
$$piint_0^3(3^2-(3-x)^2),dx$$
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
add a comment |Â
up vote
1
down vote
accepted
The region revolved looks like this:
^y
|
3#######
|#####
| ###
o--#--->x
3
Consider each side of the absolute value separately. For this region's left half with $0le xle3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3le xle6$, the lower bound changes to $y=x-3$.
The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation.
$$piint_0^3(3^2-(3-x)^2),dx$$
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The region revolved looks like this:
^y
|
3#######
|#####
| ###
o--#--->x
3
Consider each side of the absolute value separately. For this region's left half with $0le xle3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3le xle6$, the lower bound changes to $y=x-3$.
The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation.
$$piint_0^3(3^2-(3-x)^2),dx$$
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
The region revolved looks like this:
^y
|
3#######
|#####
| ###
o--#--->x
3
Consider each side of the absolute value separately. For this region's left half with $0le xle3$, the lower bound is $y=3-x$ and the upper bound is $y=3$. For the right half with $3le xle6$, the lower bound changes to $y=x-3$.
The integrals can then be set up for the disc method. I show the left half's integral below; the right half is for you to complete, as well as the evaluation.
$$piint_0^3(3^2-(3-x)^2),dx$$
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
answered Jul 18 at 17:56
Parcly Taxel
33.6k136588
33.6k136588
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
add a comment |Â
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
Your illustration helps incredibly. Thank you very much!
– jackbenimbo
Jul 18 at 18:08
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
To be clear then, the right region's integral would be: À * [integral from b to a of (3)^2−(x-3)^2 dx], correct? I would then proceed to add the two integrals? This is how I understand it, and would be grateful for a confirmation of thought process.
– jackbenimbo
Jul 20 at 3:50
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
@jackbenimbo $a=3$ and $b=6$ are the limits of the second integral, yes, while you do add the two parts together, yes.
– Parcly Taxel
Jul 20 at 5:26
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
Fully clear - thanks!
– jackbenimbo
Jul 20 at 5:30
add a comment |Â
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1
For the absolute value you need to consider the domain: y=|x-3| has the value x-3 if x $geq$ 3 and otherwise y has the value -(x-3)=3-x
– gd1035
Jul 18 at 17:49
This clears it up for me. Thank you @gd1035
– jackbenimbo
Jul 18 at 18:06