Find area of a curved surface within a condition

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Curved surface r given by $mathbfr=(u+v,u-v,uv)^T$. Calculate
area of a part of the curved surface that satisfies $u^2 + v^2 leq 1$

Here what I have done:

$mathbfr_u=fracpartial mathbfrpartial u= (1, 1, v)^T$

$mathbfr_v=fracpartial mathbfrpartial u= (1, 1, v)^T$

Hence
$mathbfr_u times mathbfr_v = (u + v, -u + v, -2)^T$

And
$|mathbfr_u times mathbfr_v| = sqrt(2(u^2 + v^2 + 2)$

Then the area should be

$$S = iint_D|mathbfr_u times mathbfr_v|dudv=iint_Dsqrt(2(u^2 + v^2 + 2)dudv$$

Let $u = rcostheta$, $v = rsintheta$ then $r^2 = u^2 + v^2$,
$u^2 + v^2 leq 1$ give $-1leq rleq 1$.

Thus

$$S =iint_Dsqrt(2(u^2 + v^2 + 2)dudv = int_0^2pidthetaint_-1^1sqrt(2(r^2 + 2)rdr=0 (!?)$$

Anything wrong with my calculation?

area surface-integrals

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asked Jul 15 at 1:34

nghia95

274

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1
down vote

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Curved surface r given by $mathbfr=(u+v,u-v,uv)^T$. Calculate
area of a part of the curved surface that satisfies $u^2 + v^2 leq 1$

Here what I have done:

$mathbfr_u=fracpartial mathbfrpartial u= (1, 1, v)^T$

$mathbfr_v=fracpartial mathbfrpartial u= (1, 1, v)^T$

Hence
$mathbfr_u times mathbfr_v = (u + v, -u + v, -2)^T$

And
$|mathbfr_u times mathbfr_v| = sqrt(2(u^2 + v^2 + 2)$

Then the area should be

$$S = iint_D|mathbfr_u times mathbfr_v|dudv=iint_Dsqrt(2(u^2 + v^2 + 2)dudv$$

Let $u = rcostheta$, $v = rsintheta$ then $r^2 = u^2 + v^2$,
$u^2 + v^2 leq 1$ give $-1leq rleq 1$.

Thus

$$S =iint_Dsqrt(2(u^2 + v^2 + 2)dudv = int_0^2pidthetaint_-1^1sqrt(2(r^2 + 2)rdr=0 (!?)$$

Anything wrong with my calculation?

area surface-integrals

share|cite|improve this question

asked Jul 15 at 1:34

nghia95

274

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Curved surface r given by $mathbfr=(u+v,u-v,uv)^T$. Calculate
area of a part of the curved surface that satisfies $u^2 + v^2 leq 1$

Here what I have done:

$mathbfr_u=fracpartial mathbfrpartial u= (1, 1, v)^T$

$mathbfr_v=fracpartial mathbfrpartial u= (1, 1, v)^T$

Hence
$mathbfr_u times mathbfr_v = (u + v, -u + v, -2)^T$

And
$|mathbfr_u times mathbfr_v| = sqrt(2(u^2 + v^2 + 2)$

Then the area should be

$$S = iint_D|mathbfr_u times mathbfr_v|dudv=iint_Dsqrt(2(u^2 + v^2 + 2)dudv$$

Let $u = rcostheta$, $v = rsintheta$ then $r^2 = u^2 + v^2$,
$u^2 + v^2 leq 1$ give $-1leq rleq 1$.

Thus

$$S =iint_Dsqrt(2(u^2 + v^2 + 2)dudv = int_0^2pidthetaint_-1^1sqrt(2(r^2 + 2)rdr=0 (!?)$$

Anything wrong with my calculation?

area surface-integrals

share|cite|improve this question

asked Jul 15 at 1:34

nghia95

274

Curved surface r given by $mathbfr=(u+v,u-v,uv)^T$. Calculate
area of a part of the curved surface that satisfies $u^2 + v^2 leq 1$

Here what I have done:

$mathbfr_u=fracpartial mathbfrpartial u= (1, 1, v)^T$

$mathbfr_v=fracpartial mathbfrpartial u= (1, 1, v)^T$

Hence
$mathbfr_u times mathbfr_v = (u + v, -u + v, -2)^T$

And
$|mathbfr_u times mathbfr_v| = sqrt(2(u^2 + v^2 + 2)$

Then the area should be

$$S = iint_D|mathbfr_u times mathbfr_v|dudv=iint_Dsqrt(2(u^2 + v^2 + 2)dudv$$

Let $u = rcostheta$, $v = rsintheta$ then $r^2 = u^2 + v^2$,
$u^2 + v^2 leq 1$ give $-1leq rleq 1$.

Thus

$$S =iint_Dsqrt(2(u^2 + v^2 + 2)dudv = int_0^2pidthetaint_-1^1sqrt(2(r^2 + 2)rdr=0 (!?)$$

Anything wrong with my calculation?

area surface-integrals

share|cite|improve this question

asked Jul 15 at 1:34

nghia95

274

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 1:34

nghia95

274

asked Jul 15 at 1:34

nghia95

274

asked Jul 15 at 1:34

nghia95

274

274

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
0
down vote



accepted

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=int_0^1int_0^2pirsqrt2(r^2+2),dtheta,dr$$
$$=int_0^12pi rsqrt2(r^2+2),dr$$
$$=fracpi2int_0^14rsqrt2(r^2+2),dr$$
$$=fracpi2left[frac23(2(r^2+2))^3/2right]_0^1=pileft(2sqrt6-frac83right)=7.013dots$$

share|cite|improve this answer

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why? Because r have to be positive?
  – nghia95
  Jul 15 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nghia95 Yes, this is polar coordinates!
  – Parcly Taxel
  Jul 15 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh really? I didn't know that. Thank you so much!
  – nghia95
  Jul 15 at 12:45
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=int_0^1int_0^2pirsqrt2(r^2+2),dtheta,dr$$
$$=int_0^12pi rsqrt2(r^2+2),dr$$
$$=fracpi2int_0^14rsqrt2(r^2+2),dr$$
$$=fracpi2left[frac23(2(r^2+2))^3/2right]_0^1=pileft(2sqrt6-frac83right)=7.013dots$$

share|cite|improve this answer

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why? Because r have to be positive?
  – nghia95
  Jul 15 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nghia95 Yes, this is polar coordinates!
  – Parcly Taxel
  Jul 15 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh really? I didn't know that. Thank you so much!
  – nghia95
  Jul 15 at 12:45
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=int_0^1int_0^2pirsqrt2(r^2+2),dtheta,dr$$
$$=int_0^12pi rsqrt2(r^2+2),dr$$
$$=fracpi2int_0^14rsqrt2(r^2+2),dr$$
$$=fracpi2left[frac23(2(r^2+2))^3/2right]_0^1=pileft(2sqrt6-frac83right)=7.013dots$$

share|cite|improve this answer

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why? Because r have to be positive?
  – nghia95
  Jul 15 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nghia95 Yes, this is polar coordinates!
  – Parcly Taxel
  Jul 15 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh really? I didn't know that. Thank you so much!
  – nghia95
  Jul 15 at 12:45
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=int_0^1int_0^2pirsqrt2(r^2+2),dtheta,dr$$
$$=int_0^12pi rsqrt2(r^2+2),dr$$
$$=fracpi2int_0^14rsqrt2(r^2+2),dr$$
$$=fracpi2left[frac23(2(r^2+2))^3/2right]_0^1=pileft(2sqrt6-frac83right)=7.013dots$$

share|cite|improve this answer

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

The limits for $r$ should be from 0 to 1 rather than from $-1$ to 1.
$$S=int_0^1int_0^2pirsqrt2(r^2+2),dtheta,dr$$
$$=int_0^12pi rsqrt2(r^2+2),dr$$
$$=fracpi2int_0^14rsqrt2(r^2+2),dr$$
$$=fracpi2left[frac23(2(r^2+2))^3/2right]_0^1=pileft(2sqrt6-frac83right)=7.013dots$$

share|cite|improve this answer

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 2:39

edited Jul 15 at 2:39

edited Jul 15 at 2:39

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

answered Jul 15 at 2:31

Parcly Taxel

33.6k136588

33.6k136588

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why? Because r have to be positive?
  – nghia95
  Jul 15 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nghia95 Yes, this is polar coordinates!
  – Parcly Taxel
  Jul 15 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh really? I didn't know that. Thank you so much!
  – nghia95
  Jul 15 at 12:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why? Because r have to be positive?
  – nghia95
  Jul 15 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nghia95 Yes, this is polar coordinates!
  – Parcly Taxel
  Jul 15 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh really? I didn't know that. Thank you so much!
  – nghia95
  Jul 15 at 12:45
  

  
  
  
  

Why? Because r have to be positive?
– nghia95
Jul 15 at 11:32

Why? Because r have to be positive?
– nghia95
Jul 15 at 11:32

@nghia95 Yes, this is polar coordinates!
– Parcly Taxel
Jul 15 at 12:03

@nghia95 Yes, this is polar coordinates!
– Parcly Taxel
Jul 15 at 12:03

Oh really? I didn't know that. Thank you so much!
– nghia95
Jul 15 at 12:45

Oh really? I didn't know that. Thank you so much!
– nghia95
Jul 15 at 12:45

add a comment | 

 

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