Mixing
![Full explanation on the cancellation property: $;sinbig(arcsin(x)big)=x,; forall; xin [-1,1]$](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Find surface area of $z=x+3$ with $x^2+y^2leq 1$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Find the surface area of $z=x+3$ with $(x,y,z)mid x^2+y^2leq 1$
So we first look at the projection of $phi(x,y)=(x,y,x+3)$ on $xy$
Then area element is $sqrt1+f_x^2+f_y^2=sqrt1+1^2+0^2=sqrt2$
$$sqrt2intint dx,dy$$
$x=r cos t,y=r sin t$
So
$$sqrt2int_0^1int_0^2pi r ,dt,dr=sqrt2pi$$
Is it correct? can we use Green/Stokes to solve it?
calculus multivariable-calculus surface-integrals
edited Aug 6 at 17:25
Robert Howard
1,325620
asked Aug 6 at 16:51
newhere
764310
add a comment |Â
up vote
2
down vote
favorite
Find the surface area of $z=x+3$ with $(x,y,z)mid x^2+y^2leq 1$
So we first look at the projection of $phi(x,y)=(x,y,x+3)$ on $xy$
Then area element is $sqrt1+f_x^2+f_y^2=sqrt1+1^2+0^2=sqrt2$
$$sqrt2intint dx,dy$$
$x=r cos t,y=r sin t$
So
$$sqrt2int_0^1int_0^2pi r ,dt,dr=sqrt2pi$$
Is it correct? can we use Green/Stokes to solve it?
calculus multivariable-calculus surface-integrals
edited Aug 6 at 17:25
Robert Howard
1,325620
asked Aug 6 at 16:51
newhere
764310
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the surface area of $z=x+3$ with $(x,y,z)mid x^2+y^2leq 1$
So we first look at the projection of $phi(x,y)=(x,y,x+3)$ on $xy$
Then area element is $sqrt1+f_x^2+f_y^2=sqrt1+1^2+0^2=sqrt2$
$$sqrt2intint dx,dy$$
$x=r cos t,y=r sin t$
So
$$sqrt2int_0^1int_0^2pi r ,dt,dr=sqrt2pi$$
Is it correct? can we use Green/Stokes to solve it?
calculus multivariable-calculus surface-integrals
edited Aug 6 at 17:25
Robert Howard
1,325620
asked Aug 6 at 16:51
newhere
764310
Find the surface area of $z=x+3$ with $(x,y,z)mid x^2+y^2leq 1$
So we first look at the projection of $phi(x,y)=(x,y,x+3)$ on $xy$
Then area element is $sqrt1+f_x^2+f_y^2=sqrt1+1^2+0^2=sqrt2$
$$sqrt2intint dx,dy$$
$x=r cos t,y=r sin t$
So
$$sqrt2int_0^1int_0^2pi r ,dt,dr=sqrt2pi$$
Is it correct? can we use Green/Stokes to solve it?
calculus multivariable-calculus surface-integrals
edited Aug 6 at 17:25
Robert Howard
1,325620
asked Aug 6 at 16:51
newhere
764310
edited Aug 6 at 17:25
Robert Howard
1,325620
edited Aug 6 at 17:25
Robert Howard
1,325620
edited Aug 6 at 17:25
Robert Howard
1,325620
1,325620
asked Aug 6 at 16:51
newhere
764310
asked Aug 6 at 16:51
newhere
764310
asked Aug 6 at 16:51
newhere
764310
764310
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
What you have is correct.
I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.
$(x,y,z) = r(t) = (cos t, sin t, cos t+3)\
r'(t) = (-sin t, cos t, -sin t)$
Now we need a function $F(x,y,z)$ such that $nabla times F$ is perpendicular to the plane and of magnitude 1.
$F = (0,frac sqrt22(x+z), 0)$ would suffice.
$int_0^2pi Fcdot dr = sqrt 2pi$
answered Aug 6 at 17:13
Doug M
39.3k31749
add a comment |Â
up vote
2
down vote
Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2sqrt 2$ that is
$$A=frac14 pi cdot 2 cdot 2sqrt 2=sqrt 2 pi$$
answered Aug 6 at 16:57
gimusi
65.4k73684
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you have is correct.
I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.
$(x,y,z) = r(t) = (cos t, sin t, cos t+3)\
r'(t) = (-sin t, cos t, -sin t)$
Now we need a function $F(x,y,z)$ such that $nabla times F$ is perpendicular to the plane and of magnitude 1.
$F = (0,frac sqrt22(x+z), 0)$ would suffice.
$int_0^2pi Fcdot dr = sqrt 2pi$
answered Aug 6 at 17:13
Doug M
39.3k31749
add a comment |Â
up vote
1
down vote
accepted
What you have is correct.
I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.
$(x,y,z) = r(t) = (cos t, sin t, cos t+3)\
r'(t) = (-sin t, cos t, -sin t)$
Now we need a function $F(x,y,z)$ such that $nabla times F$ is perpendicular to the plane and of magnitude 1.
$F = (0,frac sqrt22(x+z), 0)$ would suffice.
$int_0^2pi Fcdot dr = sqrt 2pi$
answered Aug 6 at 17:13
Doug M
39.3k31749
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you have is correct.
I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.
$(x,y,z) = r(t) = (cos t, sin t, cos t+3)\
r'(t) = (-sin t, cos t, -sin t)$
Now we need a function $F(x,y,z)$ such that $nabla times F$ is perpendicular to the plane and of magnitude 1.
$F = (0,frac sqrt22(x+z), 0)$ would suffice.
$int_0^2pi Fcdot dr = sqrt 2pi$
answered Aug 6 at 17:13
Doug M
39.3k31749
What you have is correct.
I am not sure why you want to use Stokes... but supposing you did...you could say that the contour of your region is bound by the contour.
$(x,y,z) = r(t) = (cos t, sin t, cos t+3)\
r'(t) = (-sin t, cos t, -sin t)$
Now we need a function $F(x,y,z)$ such that $nabla times F$ is perpendicular to the plane and of magnitude 1.
$F = (0,frac sqrt22(x+z), 0)$ would suffice.
$int_0^2pi Fcdot dr = sqrt 2pi$
answered Aug 6 at 17:13
Doug M
39.3k31749
edited Aug 7 at 6:20
answered Aug 6 at 17:13
Doug M
39.3k31749
answered Aug 6 at 17:13
Doug M
39.3k31749
answered Aug 6 at 17:13
Doug M
39.3k31749
39.3k31749
add a comment |Â
add a comment |Â
up vote
2
down vote
Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2sqrt 2$ that is
$$A=frac14 pi cdot 2 cdot 2sqrt 2=sqrt 2 pi$$
answered Aug 6 at 16:57
gimusi
65.4k73684
add a comment |Â
up vote
2
down vote
Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2sqrt 2$ that is
$$A=frac14 pi cdot 2 cdot 2sqrt 2=sqrt 2 pi$$
answered Aug 6 at 16:57
gimusi
65.4k73684
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2sqrt 2$ that is
$$A=frac14 pi cdot 2 cdot 2sqrt 2=sqrt 2 pi$$
answered Aug 6 at 16:57
gimusi
65.4k73684
Yes it is correct, indeed the intersection is the area enclosed by an ellipse with axes $a=2$ and $b=2sqrt 2$ that is
$$A=frac14 pi cdot 2 cdot 2sqrt 2=sqrt 2 pi$$
answered Aug 6 at 16:57
gimusi
65.4k73684
answered Aug 6 at 16:57
gimusi
65.4k73684
answered Aug 6 at 16:57
gimusi
65.4k73684
answered Aug 6 at 16:57
gimusi
65.4k73684
65.4k73684
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2874086%2ffind-surface-area-of-z-x3-with-x2y2-leq-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password