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Line integral for surface area

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0

$begingroup$

Use a line integral to find the area of the surface that extends upward from the semicircle $y=sqrt{4-x^2}$ in the $xy$-plane to the surface $z=3x^4y$.

I know how to compute line integrals but I'm unsure about how to use them to find surface areas. Any help would be great. Thank you in advance!

multivariable-calculus line-integrals

share|cite|improve this question

asked Mar 20 at 18:08

James DoneJames Done

808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The line integral is the area between a surface and the line you are integrating over. Think of how a normal integral is the area under the curve. A line integral is the surface area of the sheet formed by connecting the line vertically to the surface, the 'area' under the curve. I.e. evaluate the line integral and you have your answer.
  $endgroup$
  – Kyle C
  Mar 20 at 18:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That makes sense, @KyleC, but I'm unsure of how to set it up.
  $endgroup$
  – James Done
  Mar 20 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Compute the line integral $int_S 3x^4y ds$ where $S$ is the semicircle you have defined.
  $endgroup$
  – Kyle C
  Mar 20 at 18:40
  

  
  
  
  

add a comment | 

0

$begingroup$

Use a line integral to find the area of the surface that extends upward from the semicircle $y=sqrt{4-x^2}$ in the $xy$-plane to the surface $z=3x^4y$.

I know how to compute line integrals but I'm unsure about how to use them to find surface areas. Any help would be great. Thank you in advance!

multivariable-calculus line-integrals

share|cite|improve this question

asked Mar 20 at 18:08

James DoneJames Done

808

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The line integral is the area between a surface and the line you are integrating over. Think of how a normal integral is the area under the curve. A line integral is the surface area of the sheet formed by connecting the line vertically to the surface, the 'area' under the curve. I.e. evaluate the line integral and you have your answer.
  $endgroup$
  – Kyle C
  Mar 20 at 18:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That makes sense, @KyleC, but I'm unsure of how to set it up.
  $endgroup$
  – James Done
  Mar 20 at 18:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Compute the line integral $int_S 3x^4y ds$ where $S$ is the semicircle you have defined.
  $endgroup$
  – Kyle C
  Mar 20 at 18:40
  

  
  
  
  

add a comment | 

$begingroup$

Use a line integral to find the area of the surface that extends upward from the semicircle $y=sqrt{4-x^2}$ in the $xy$-plane to the surface $z=3x^4y$.

I know how to compute line integrals but I'm unsure about how to use them to find surface areas. Any help would be great. Thank you in advance!

multivariable-calculus line-integrals

share|cite|improve this question

asked Mar 20 at 18:08

James DoneJames Done

808

$endgroup$

share|cite|improve this question

asked Mar 20 at 18:08

James DoneJames Done

808

share|cite|improve this question

asked Mar 20 at 18:08

James DoneJames Done

808

asked Mar 20 at 18:08

James DoneJames Done

808

asked Mar 20 at 18:08

James DoneJames Done

808

1 Answer
1

active

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1

$begingroup$

The appropriate integral here, for the area $A$ of the vertical surface between a curve $C$ and the surface $z=f(x,y)$, is $int_C f(x,y),ds$. What is that $ds$? Arclength; when we parametrize the curve as $(x(t),y(t))$, $ds$ becomes $sqrt{dx^2+dy^2}=sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2},dt$.

So now, the next step we need is a parametrization of the semicircle. The obvious place to start is to use $x$ as the parameter: $(x,y)=(t,sqrt{4-t^2})$. The $x$-coordinate ranges between $-2$ and $2$, so
$$A = int_{-2}^2 3t^4sqrt{4-t^2}cdotsqrt{1+left(frac{-t}{sqrt{4-t^2}}right)^2},dt$$
Alternatively, it's part of a circle of radius $2$. We could use an angular parametrization $(x,y)=(2costheta,2sintheta)$. Our semicircle is the half with $theta$ positive, ranging from $0$ to $pi$, so
$$A = int_0^{pi} 3(2costheta)^4cdot 2sinthetacdotsqrt{(-2sintheta)^2+(2costheta)^2},dtheta$$
Both forms could use further simplification, of course. Take your pick on which one you'd rather work with.

share|cite|improve this answer

answered Mar 20 at 18:42

jmerryjmerry

17k11633

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add a comment | 

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1

$begingroup$

The appropriate integral here, for the area $A$ of the vertical surface between a curve $C$ and the surface $z=f(x,y)$, is $int_C f(x,y),ds$. What is that $ds$? Arclength; when we parametrize the curve as $(x(t),y(t))$, $ds$ becomes $sqrt{dx^2+dy^2}=sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2},dt$.

So now, the next step we need is a parametrization of the semicircle. The obvious place to start is to use $x$ as the parameter: $(x,y)=(t,sqrt{4-t^2})$. The $x$-coordinate ranges between $-2$ and $2$, so
$$A = int_{-2}^2 3t^4sqrt{4-t^2}cdotsqrt{1+left(frac{-t}{sqrt{4-t^2}}right)^2},dt$$
Alternatively, it's part of a circle of radius $2$. We could use an angular parametrization $(x,y)=(2costheta,2sintheta)$. Our semicircle is the half with $theta$ positive, ranging from $0$ to $pi$, so
$$A = int_0^{pi} 3(2costheta)^4cdot 2sinthetacdotsqrt{(-2sintheta)^2+(2costheta)^2},dtheta$$
Both forms could use further simplification, of course. Take your pick on which one you'd rather work with.

share|cite|improve this answer

answered Mar 20 at 18:42

jmerryjmerry

17k11633

$endgroup$

add a comment | 

1

$begingroup$

The appropriate integral here, for the area $A$ of the vertical surface between a curve $C$ and the surface $z=f(x,y)$, is $int_C f(x,y),ds$. What is that $ds$? Arclength; when we parametrize the curve as $(x(t),y(t))$, $ds$ becomes $sqrt{dx^2+dy^2}=sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2},dt$.

So now, the next step we need is a parametrization of the semicircle. The obvious place to start is to use $x$ as the parameter: $(x,y)=(t,sqrt{4-t^2})$. The $x$-coordinate ranges between $-2$ and $2$, so
$$A = int_{-2}^2 3t^4sqrt{4-t^2}cdotsqrt{1+left(frac{-t}{sqrt{4-t^2}}right)^2},dt$$
Alternatively, it's part of a circle of radius $2$. We could use an angular parametrization $(x,y)=(2costheta,2sintheta)$. Our semicircle is the half with $theta$ positive, ranging from $0$ to $pi$, so
$$A = int_0^{pi} 3(2costheta)^4cdot 2sinthetacdotsqrt{(-2sintheta)^2+(2costheta)^2},dtheta$$
Both forms could use further simplification, of course. Take your pick on which one you'd rather work with.

share|cite|improve this answer

answered Mar 20 at 18:42

jmerryjmerry

17k11633

$endgroup$

add a comment | 

$begingroup$

The appropriate integral here, for the area $A$ of the vertical surface between a curve $C$ and the surface $z=f(x,y)$, is $int_C f(x,y),ds$. What is that $ds$? Arclength; when we parametrize the curve as $(x(t),y(t))$, $ds$ becomes $sqrt{dx^2+dy^2}=sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2},dt$.

So now, the next step we need is a parametrization of the semicircle. The obvious place to start is to use $x$ as the parameter: $(x,y)=(t,sqrt{4-t^2})$. The $x$-coordinate ranges between $-2$ and $2$, so
$$A = int_{-2}^2 3t^4sqrt{4-t^2}cdotsqrt{1+left(frac{-t}{sqrt{4-t^2}}right)^2},dt$$
Alternatively, it's part of a circle of radius $2$. We could use an angular parametrization $(x,y)=(2costheta,2sintheta)$. Our semicircle is the half with $theta$ positive, ranging from $0$ to $pi$, so
$$A = int_0^{pi} 3(2costheta)^4cdot 2sinthetacdotsqrt{(-2sintheta)^2+(2costheta)^2},dtheta$$
Both forms could use further simplification, of course. Take your pick on which one you'd rather work with.

share|cite|improve this answer

answered Mar 20 at 18:42

jmerryjmerry

17k11633

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:42

jmerryjmerry

17k11633

answered Mar 20 at 18:42

jmerryjmerry

17k11633

answered Mar 20 at 18:42

jmerryjmerry

17k11633