Mixing
surface area with integrals
$begingroup$
I'm working on a problem in my textbook and am confused on how to set up the integral.
"Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.
I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.
integration multivariable-calculus surface-integrals
edited Jan 8 at 21:47
user376343
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
$endgroup$
add a comment |
$begingroup$
I'm working on a problem in my textbook and am confused on how to set up the integral.
"Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.
I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.
integration multivariable-calculus surface-integrals
edited Jan 8 at 21:47
user376343
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
$endgroup$
add a comment |
$begingroup$
I'm working on a problem in my textbook and am confused on how to set up the integral.
"Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.
I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.
integration multivariable-calculus surface-integrals
edited Jan 8 at 21:47
user376343
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
$endgroup$
I'm working on a problem in my textbook and am confused on how to set up the integral.
"Find the surface area of the part of the hyperbolic paraboloid $z= x^2 - y^2$ that lies in the first octant and is inside the cylinder $x^2 +y^2 = 1$.
I have figured out the equation to integrate: $sqrt{4x^2+4y^2+1}$ from finding the partial derivatives and taking the magnitude of the cross product.
integration multivariable-calculus surface-integrals
integration multivariable-calculus surface-integrals
edited Jan 8 at 21:47
user376343
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
edited Jan 8 at 21:47
user376343
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
edited Jan 8 at 21:47
user376343
3,4233826
edited Jan 8 at 21:47
user376343
3,4233826
edited Jan 8 at 21:47
user376343
3,4233826
3,4233826
asked Jan 8 at 21:43
marbeiiimarbeiii
113
asked Jan 8 at 21:43
marbeiiimarbeiii
113
asked Jan 8 at 21:43
marbeiiimarbeiii
113
113
add a comment |
add a comment |
1 Answer
1
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$begingroup$
You need to parameterize the integration region. One possible choice is to take
$$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
$$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$
Then the surface element is
$$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
The area of the region of interest (call it $R$) is given by the surface integral,
$$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$
answered Jan 8 at 21:58
user170231user170231
4,04711429
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You need to parameterize the integration region. One possible choice is to take
$$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
$$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$
Then the surface element is
$$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
The area of the region of interest (call it $R$) is given by the surface integral,
$$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$
answered Jan 8 at 21:58
user170231user170231
4,04711429
$endgroup$
add a comment |
$begingroup$
You need to parameterize the integration region. One possible choice is to take
$$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
$$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$
Then the surface element is
$$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
The area of the region of interest (call it $R$) is given by the surface integral,
$$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$
answered Jan 8 at 21:58
user170231user170231
4,04711429
$endgroup$
add a comment |
$begingroup$
You need to parameterize the integration region. One possible choice is to take
$$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
$$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$
Then the surface element is
$$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
The area of the region of interest (call it $R$) is given by the surface integral,
$$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$
answered Jan 8 at 21:58
user170231user170231
4,04711429
$endgroup$
You need to parameterize the integration region. One possible choice is to take
$$mathbf s(u,v)=(x(u,v),y(u,v),z(u,v))=(ucos v,usin v,u^2cos2v)$$
with $0le ule1$ and $0le vle2pi$. We took a cue to use polar/cylindrical coordinates from the cylinder's equation, $x^2+y^2=1$. Then the $z$ coordinate is determined by the equation of the surface,
$$z=x^2-y^2=u^2cos^2v-u^2sin^2v=u^2cos2v$$
Then the surface element is
$$mathrm dS=left|frac{partialmathbf s}{partial u}timesfrac{partialmathbf s}{partial v}right|,mathrm du,mathrm dv=usqrt{1+4u^2},mathrm du,mathrm dv$$
The area of the region of interest (call it $R$) is given by the surface integral,
$$iint_Rmathrm dS=int_{v=0}^{v=2pi}int_{u=0}^{u=1}usqrt{1+4u^2},mathrm du,mathrm dv=frac{(5sqrt5-1)pi}6$$
answered Jan 8 at 21:58
user170231user170231
4,04711429
answered Jan 8 at 21:58
user170231user170231
4,04711429
answered Jan 8 at 21:58
user170231user170231
4,04711429
answered Jan 8 at 21:58
user170231user170231
4,04711429
4,04711429
add a comment |
add a comment |
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