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Show that the solid generated by the revolution of the region has finite volume but infinite surface area
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Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
asked 3 hours ago
md emon
17111
add a comment |Â
up vote
3
down vote
favorite
Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
asked 3 hours ago
md emon
17111
1
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
1
See also here
– Jyrki Lahtonen
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
asked 3 hours ago
md emon
17111
Q:Let R be the region to the right of $x=1$ that is bounded by the $x$-axis and the curve $y=frac1x$. Show that the solid generated bt the revolution of the region about the $x$-axis has finite volume but infinite surface area.
My Approach:$$
V = pi int_1^infty frac1x^2 dx=pilim_t to inftyint_1^tfrac1x^2 dx=pi
$$Hence i can said that the revolution of the region has finite volume.But i haven't a very good idea about improper integral then how could I prove$$A=2piint_1^infty frac1xsqrt1+frac1x^4 dx=divergent$$Any hints or solution will be appreciated.
Thanks in advance.
calculus solid-of-revolution
calculus solid-of-revolution
asked 3 hours ago
md emon
17111
asked 3 hours ago
md emon
17111
edited 2 hours ago
asked 3 hours ago
md emon
17111
asked 3 hours ago
md emon
17111
asked 3 hours ago
md emon
17111
17111
1
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
1
See also here
– Jyrki Lahtonen
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago
add a comment |Â
1
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
1
See also here
– Jyrki Lahtonen
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago
1
1
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
1
1
See also here
– Jyrki Lahtonen
2 hours ago
See also here
– Jyrki Lahtonen
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$
answered 3 hours ago
Andrei
8,4632923
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
add a comment |Â
up vote
2
down vote
Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.
answered 3 hours ago
Parcly Taxel
34k136890
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$
answered 3 hours ago
Andrei
8,4632923
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
add a comment |Â
up vote
3
down vote
For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$
answered 3 hours ago
Andrei
8,4632923
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$
answered 3 hours ago
Andrei
8,4632923
For positive $x$, $sqrt1+x^-4$ is always greater than $1$. Then $$A>2piint_1^infty frac1xdx=2pi(lninfty-ln 1)=infty$$
answered 3 hours ago
Andrei
8,4632923
answered 3 hours ago
Andrei
8,4632923
answered 3 hours ago
Andrei
8,4632923
answered 3 hours ago
Andrei
8,4632923
8,4632923
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
add a comment |Â
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Thanks @Andrei. I really suffer with improper integral to show divergent or convergent ':( can you suggest me how to make a good concept about comparison test
– md emon
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
Good concept :.... It depends on the problem. The idea is that if you can solve a related but simpler integral that converges/diverges and compare your desired function with that, without calculating the integral. Suppose my simplified function $f_s$ is always positive in a certain interval $a$ to $b$. Then if $f(x)>f_s(x)$ you get $$int_a^b f(x)dx>int_a^b f_s(x)dx$$
– Andrei
2 hours ago
add a comment |Â
up vote
2
down vote
Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.
answered 3 hours ago
Parcly Taxel
34k136890
add a comment |Â
up vote
2
down vote
Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.
answered 3 hours ago
Parcly Taxel
34k136890
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.
answered 3 hours ago
Parcly Taxel
34k136890
Lower-bound the surface area integral with another divergent integral. $sqrt1+1/x^4>1$ for all $xge 1$, so the surface area integral is greater than
$$int_1^inftyfrac 1x,dx$$
which diverges.
answered 3 hours ago
Parcly Taxel
34k136890
answered 3 hours ago
Parcly Taxel
34k136890
answered 3 hours ago
Parcly Taxel
34k136890
answered 3 hours ago
Parcly Taxel
34k136890
34k136890
add a comment |Â
add a comment |Â
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1
Are you sure you got the right curve? Looks more like the surface gotten by revolving $y=1/x$ instead of $y=1/x^2$. Anyway, the judgement day is here.
– Jyrki Lahtonen
3 hours ago
1
See also here
– Jyrki Lahtonen
2 hours ago
thanks @Jyrki Lahtonen. Yeah i did type mistake :)
– md emon
2 hours ago