Surface integral for area calculation
Is my procedure correct?
Calculate paraboloid area portion of equations
$$P equiv (u cos v, u sin v, u^2) $$
with $0 leq v leq dfrac{pi}{4}$ and $0 leq u leq dfrac{1}{2}tan v$
This is my attempt:
$bullet$ vector first derivatives:
$$P'_u = (cos v, sin v, 2u)$$
$$P'_v = (-usin v, ucos v, 0)$$
$bullet$ cross product between $P'_u$ and $P'_v$
$$P'_u wedge P'_v = (-2u^2cos v, 2u^2sin v, u)$$
$bullet$ cross product module
$$N(u,v) = usqrt{4u^2 + 1} $$
$bullet$ double integral
$$displaystyle iint_P usqrt{4u^2 + 1} ; du ; dv = int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du$$
where $alpha(v) = 0$ and $beta(v) = dfrac{1}{2} tan dfrac{pi}{4} = dfrac{1}{2}$
$$displaystyle int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du = displaystyle dfrac{1}{12} (4u^2 + 1)^{3/2} Big|^{u=1/2}_{u=0} = dfrac{1}{12}(2sqrt{2} - 1)$$
Thank you in advance
surface-integrals
asked Nov 25 '18 at 17:21
user3204810
1877
add a comment |
Is my procedure correct?
Calculate paraboloid area portion of equations
$$P equiv (u cos v, u sin v, u^2) $$
with $0 leq v leq dfrac{pi}{4}$ and $0 leq u leq dfrac{1}{2}tan v$
This is my attempt:
$bullet$ vector first derivatives:
$$P'_u = (cos v, sin v, 2u)$$
$$P'_v = (-usin v, ucos v, 0)$$
$bullet$ cross product between $P'_u$ and $P'_v$
$$P'_u wedge P'_v = (-2u^2cos v, 2u^2sin v, u)$$
$bullet$ cross product module
$$N(u,v) = usqrt{4u^2 + 1} $$
$bullet$ double integral
$$displaystyle iint_P usqrt{4u^2 + 1} ; du ; dv = int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du$$
where $alpha(v) = 0$ and $beta(v) = dfrac{1}{2} tan dfrac{pi}{4} = dfrac{1}{2}$
$$displaystyle int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du = displaystyle dfrac{1}{12} (4u^2 + 1)^{3/2} Big|^{u=1/2}_{u=0} = dfrac{1}{12}(2sqrt{2} - 1)$$
Thank you in advance
surface-integrals
asked Nov 25 '18 at 17:21
user3204810
1877
add a comment |
Is my procedure correct?
Calculate paraboloid area portion of equations
$$P equiv (u cos v, u sin v, u^2) $$
with $0 leq v leq dfrac{pi}{4}$ and $0 leq u leq dfrac{1}{2}tan v$
This is my attempt:
$bullet$ vector first derivatives:
$$P'_u = (cos v, sin v, 2u)$$
$$P'_v = (-usin v, ucos v, 0)$$
$bullet$ cross product between $P'_u$ and $P'_v$
$$P'_u wedge P'_v = (-2u^2cos v, 2u^2sin v, u)$$
$bullet$ cross product module
$$N(u,v) = usqrt{4u^2 + 1} $$
$bullet$ double integral
$$displaystyle iint_P usqrt{4u^2 + 1} ; du ; dv = int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du$$
where $alpha(v) = 0$ and $beta(v) = dfrac{1}{2} tan dfrac{pi}{4} = dfrac{1}{2}$
$$displaystyle int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du = displaystyle dfrac{1}{12} (4u^2 + 1)^{3/2} Big|^{u=1/2}_{u=0} = dfrac{1}{12}(2sqrt{2} - 1)$$
Thank you in advance
surface-integrals
asked Nov 25 '18 at 17:21
user3204810
1877
Is my procedure correct?
Calculate paraboloid area portion of equations
$$P equiv (u cos v, u sin v, u^2) $$
with $0 leq v leq dfrac{pi}{4}$ and $0 leq u leq dfrac{1}{2}tan v$
This is my attempt:
$bullet$ vector first derivatives:
$$P'_u = (cos v, sin v, 2u)$$
$$P'_v = (-usin v, ucos v, 0)$$
$bullet$ cross product between $P'_u$ and $P'_v$
$$P'_u wedge P'_v = (-2u^2cos v, 2u^2sin v, u)$$
$bullet$ cross product module
$$N(u,v) = usqrt{4u^2 + 1} $$
$bullet$ double integral
$$displaystyle iint_P usqrt{4u^2 + 1} ; du ; dv = int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du$$
where $alpha(v) = 0$ and $beta(v) = dfrac{1}{2} tan dfrac{pi}{4} = dfrac{1}{2}$
$$displaystyle int^{beta(v)}_{alpha(v)} usqrt{4u^2 + 1} ; du = displaystyle dfrac{1}{12} (4u^2 + 1)^{3/2} Big|^{u=1/2}_{u=0} = dfrac{1}{12}(2sqrt{2} - 1)$$
Thank you in advance
surface-integrals
surface-integrals
asked Nov 25 '18 at 17:21
user3204810
1877
asked Nov 25 '18 at 17:21
user3204810
1877
edited Nov 25 '18 at 18:41
asked Nov 25 '18 at 17:21
user3204810
1877
asked Nov 25 '18 at 17:21
user3204810
1877
asked Nov 25 '18 at 17:21
user3204810
1877
1877
add a comment |
add a comment |
1 Answer
1
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oldest
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The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have
begin{align}
int_P , mathrm dA &= int_{v=0}^{pi/4} int_{u=0}^{(tan v)/2} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
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
The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have
begin{align}
int_P , mathrm dA &= int_{v=0}^{pi/4} int_{u=0}^{(tan v)/2} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
add a comment |
The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have
begin{align}
int_P , mathrm dA &= int_{v=0}^{pi/4} int_{u=0}^{(tan v)/2} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
add a comment |
The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have
begin{align}
int_P , mathrm dA &= int_{v=0}^{pi/4} int_{u=0}^{(tan v)/2} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
The partial derivative $P_u'$ shouldn't have a minus sign in the second component. Otherwise, everything was fine until you inserted everything into the integral. Notice that your bound for $u$ depends on $v$, so you should have
begin{align}
int_P , mathrm dA &= int_{v=0}^{pi/4} int_{u=0}^{(tan v)/2} usqrt{4u^2+1} ,mathrm du , mathrm dv\
&= frac{1}{12}int_{v=0}^{pi/4}left((tan^2v+1)^{3/2}- 1right) , mathrm dv \
&= frac{1}{12}int_{v=0}^{pi/4} sec^3 v , mathrm dv - frac{pi}{48},
end{align}
etc.
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
edited Nov 25 '18 at 17:40
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
answered Nov 25 '18 at 17:30
MisterRiemann
5,7791624
5,7791624
add a comment |
add a comment |
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