Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about x = 6.
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3
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y = x, y = 0, x = 3.
Revolving around x = 6, what is the volume of the shape formed using shell method?
I've tried this:
http://www3.wolframalpha.com/Calculate/MSP/MSP447521i89ce0d69c7cb5000040fc75g3e5g51hca?MSPStoreType=image/gif&s=44
From what I've researched online and from my notes in class, this should be working but I can't find what I'm doing wrong.
I've also tried integrating from 3 to 6. Not sure what else I could be doing wrong. Thank you.
calculus
asked Sep 3 at 1:33
Ivan
234
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up vote
3
down vote
favorite
y = x, y = 0, x = 3.
Revolving around x = 6, what is the volume of the shape formed using shell method?
I've tried this:
http://www3.wolframalpha.com/Calculate/MSP/MSP447521i89ce0d69c7cb5000040fc75g3e5g51hca?MSPStoreType=image/gif&s=44
From what I've researched online and from my notes in class, this should be working but I can't find what I'm doing wrong.
I've also tried integrating from 3 to 6. Not sure what else I could be doing wrong. Thank you.
calculus
asked Sep 3 at 1:33
Ivan
234
1
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
y = x, y = 0, x = 3.
Revolving around x = 6, what is the volume of the shape formed using shell method?
I've tried this:
http://www3.wolframalpha.com/Calculate/MSP/MSP447521i89ce0d69c7cb5000040fc75g3e5g51hca?MSPStoreType=image/gif&s=44
From what I've researched online and from my notes in class, this should be working but I can't find what I'm doing wrong.
I've also tried integrating from 3 to 6. Not sure what else I could be doing wrong. Thank you.
calculus
asked Sep 3 at 1:33
Ivan
234
y = x, y = 0, x = 3.
Revolving around x = 6, what is the volume of the shape formed using shell method?
I've tried this:
http://www3.wolframalpha.com/Calculate/MSP/MSP447521i89ce0d69c7cb5000040fc75g3e5g51hca?MSPStoreType=image/gif&s=44
From what I've researched online and from my notes in class, this should be working but I can't find what I'm doing wrong.
I've also tried integrating from 3 to 6. Not sure what else I could be doing wrong. Thank you.
calculus
calculus
asked Sep 3 at 1:33
Ivan
234
asked Sep 3 at 1:33
Ivan
234
edited Sep 3 at 1:42
asked Sep 3 at 1:33
Ivan
234
asked Sep 3 at 1:33
Ivan
234
asked Sep 3 at 1:33
Ivan
234
234
1
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36
add a comment |Â
1
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36
1
1
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.
answered Sep 3 at 1:53
rogerl
16.7k22745
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up vote
4
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$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$
answered Sep 3 at 1:52
Chris Custer
6,9222622
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.
answered Sep 3 at 1:53
rogerl
16.7k22745
add a comment |Â
up vote
4
down vote
accepted
At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.
answered Sep 3 at 1:53
rogerl
16.7k22745
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.
answered Sep 3 at 1:53
rogerl
16.7k22745
At any $x$, you have a shell whose radius is $6-x$ and whose height is $x$, so its volume is $2pi x(6-x)Delta x$.
answered Sep 3 at 1:53
rogerl
16.7k22745
answered Sep 3 at 1:53
rogerl
16.7k22745
answered Sep 3 at 1:53
rogerl
16.7k22745
answered Sep 3 at 1:53
rogerl
16.7k22745
16.7k22745
add a comment |Â
add a comment |Â
up vote
4
down vote
$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$
answered Sep 3 at 1:52
Chris Custer
6,9222622
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
add a comment |Â
up vote
4
down vote
$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$
answered Sep 3 at 1:52
Chris Custer
6,9222622
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$
answered Sep 3 at 1:52
Chris Custer
6,9222622
$2piint_0^3(6-x)(x-0)mathbb dx=2piint_0^3(6x-x^2)mathbb dx=2pi [3x^2-fracx^33]_0^3=2pi[18-0]=36pi$
answered Sep 3 at 1:52
Chris Custer
6,9222622
answered Sep 3 at 1:52
Chris Custer
6,9222622
answered Sep 3 at 1:52
Chris Custer
6,9222622
answered Sep 3 at 1:52
Chris Custer
6,9222622
6,9222622
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
add a comment |Â
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Thank you so much. I wasn't thinking about it in the sense of subtracting but rather adding the extra length so I was stuck with adding 3.
– Ivan
Sep 3 at 1:59
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
Sure. The radius is $(6-x)$, when rotating about $x=6$...
– Chris Custer
Sep 3 at 2:06
add a comment |Â
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1
please include your thoughts about questions or what You have treid
– Deepesh Meena
Sep 3 at 1:36