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How to calculate the surface area of parametric surface?
up vote
3
down vote
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Suppose you have the surface $z=3xy$ and you want to find the area that lies within the cylinder $x^2+y^2leq 1$.
My homework is forcing me to use the parameterization
$$textbf{r}_1(s,t)= <scos(t), ssin(t), 3s^2sin(t)cos(t)>$$
I am having a difficult time visualizing this parameterization, and I do not have any graphing software to graph the surface, but I want to make sure I understand this concept.
This is quite obvious, but I want to be sure; using the above parameterization, I am not parameterizing the entire surface, right? If I wanted to, I assume the parameterization would be $$textbf{r}_2(s,t) = <s,t,3st>$$
Instead, is $textbf{r}_1$ just the parameterization adjusted for the region - the region being the cylinder $x^2+y^2leq 1$? That is, are we just making a revolution around $z=3xy$?
Any insight would be helpful.
multivariable-calculus vectors surfaces parametrization
asked 19 hours ago
Art
3037
add a comment |
up vote
3
down vote
favorite
Suppose you have the surface $z=3xy$ and you want to find the area that lies within the cylinder $x^2+y^2leq 1$.
My homework is forcing me to use the parameterization
$$textbf{r}_1(s,t)= <scos(t), ssin(t), 3s^2sin(t)cos(t)>$$
I am having a difficult time visualizing this parameterization, and I do not have any graphing software to graph the surface, but I want to make sure I understand this concept.
This is quite obvious, but I want to be sure; using the above parameterization, I am not parameterizing the entire surface, right? If I wanted to, I assume the parameterization would be $$textbf{r}_2(s,t) = <s,t,3st>$$
Instead, is $textbf{r}_1$ just the parameterization adjusted for the region - the region being the cylinder $x^2+y^2leq 1$? That is, are we just making a revolution around $z=3xy$?
Any insight would be helpful.
multivariable-calculus vectors surfaces parametrization
asked 19 hours ago
Art
3037
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose you have the surface $z=3xy$ and you want to find the area that lies within the cylinder $x^2+y^2leq 1$.
My homework is forcing me to use the parameterization
$$textbf{r}_1(s,t)= <scos(t), ssin(t), 3s^2sin(t)cos(t)>$$
I am having a difficult time visualizing this parameterization, and I do not have any graphing software to graph the surface, but I want to make sure I understand this concept.
This is quite obvious, but I want to be sure; using the above parameterization, I am not parameterizing the entire surface, right? If I wanted to, I assume the parameterization would be $$textbf{r}_2(s,t) = <s,t,3st>$$
Instead, is $textbf{r}_1$ just the parameterization adjusted for the region - the region being the cylinder $x^2+y^2leq 1$? That is, are we just making a revolution around $z=3xy$?
Any insight would be helpful.
multivariable-calculus vectors surfaces parametrization
asked 19 hours ago
Art
3037
Suppose you have the surface $z=3xy$ and you want to find the area that lies within the cylinder $x^2+y^2leq 1$.
My homework is forcing me to use the parameterization
$$textbf{r}_1(s,t)= <scos(t), ssin(t), 3s^2sin(t)cos(t)>$$
I am having a difficult time visualizing this parameterization, and I do not have any graphing software to graph the surface, but I want to make sure I understand this concept.
This is quite obvious, but I want to be sure; using the above parameterization, I am not parameterizing the entire surface, right? If I wanted to, I assume the parameterization would be $$textbf{r}_2(s,t) = <s,t,3st>$$
Instead, is $textbf{r}_1$ just the parameterization adjusted for the region - the region being the cylinder $x^2+y^2leq 1$? That is, are we just making a revolution around $z=3xy$?
Any insight would be helpful.
multivariable-calculus vectors surfaces parametrization
multivariable-calculus vectors surfaces parametrization
asked 19 hours ago
Art
3037
asked 19 hours ago
Art
3037
asked 19 hours ago
Art
3037
asked 19 hours ago
Art
3037
asked 19 hours ago
Art
3037
3037
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.
$mathbf r_1$ is the cylindrical parameterization, which, generally, is
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$
In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$
I got to this integral by the following way:
$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$
answered 18 hours ago
Lorenzo B.
1,6222419
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
add a comment |
up vote
4
down vote
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.
answered 18 hours ago
Christian Blatter
170k7111324
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.
$mathbf r_1$ is the cylindrical parameterization, which, generally, is
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$
In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$
I got to this integral by the following way:
$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$
answered 18 hours ago
Lorenzo B.
1,6222419
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
add a comment |
up vote
1
down vote
accepted
You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.
$mathbf r_1$ is the cylindrical parameterization, which, generally, is
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$
In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$
I got to this integral by the following way:
$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$
answered 18 hours ago
Lorenzo B.
1,6222419
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.
$mathbf r_1$ is the cylindrical parameterization, which, generally, is
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$
In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$
I got to this integral by the following way:
$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$
answered 18 hours ago
Lorenzo B.
1,6222419
You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.
$mathbf r_1$ is the cylindrical parameterization, which, generally, is
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$
In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$
I got to this integral by the following way:
$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$
answered 18 hours ago
Lorenzo B.
1,6222419
edited 14 hours ago
answered 18 hours ago
Lorenzo B.
1,6222419
answered 18 hours ago
Lorenzo B.
1,6222419
answered 18 hours ago
Lorenzo B.
1,6222419
1,6222419
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
add a comment |
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0>$? We have a real "floppy disc" here!
– Christian Blatter
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $oint_{x^2+y^2le1}3xymathop{mathrm dxmathrm dy}=3int_{-1}^1xmathop{mathrm dx} int_{-sqrt{1-x^2}}^{sqrt{1-x^2}}ymathop{mathrm dy}=0$. I honestly do not see where the mistake is
– Lorenzo B.
17 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
I understand your answer (+1). Why mine is wrong?
– Lorenzo B.
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
What made you arrive at the integrand $3s^3cos tsin t>$? – That's my last word on this matter.
– Christian Blatter
14 hours ago
add a comment |
up vote
4
down vote
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.
answered 18 hours ago
Christian Blatter
170k7111324
add a comment |
up vote
4
down vote
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.
answered 18 hours ago
Christian Blatter
170k7111324
add a comment |
up vote
4
down vote
up vote
4
down vote
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.
answered 18 hours ago
Christian Blatter
170k7111324
Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.
answered 18 hours ago
Christian Blatter
170k7111324
edited 15 hours ago
answered 18 hours ago
Christian Blatter
170k7111324
answered 18 hours ago
Christian Blatter
170k7111324
answered 18 hours ago
Christian Blatter
170k7111324
170k7111324
add a comment |
add a comment |
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