Area Under a Parametric Curve (did I do it correctly?)

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Homework Help Overview

The discussion revolves around finding the area under a parametric curve defined by the equations x=5sin(t) and y=2cos(t), with the domain of t being from 0 to 2π. The original poster initially interprets the curve as a circle but later acknowledges it as an ellipse.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the formula for the area under a parametric curve, specifically questioning the integration of g(t)h'(t) and its relevance to the ellipse. There is also exploration of the implications of integrating over the full domain versus a quarter of the ellipse.

Discussion Status

Participants are actively engaging with the problem, clarifying the nature of the curve and the appropriate integration limits. Some guidance is provided regarding the area formula for an ellipse, and there is recognition of the need to consider the entire ellipse rather than just a quarter.

Contextual Notes

There is mention of confusion regarding the integration limits and the relationship between the parametric equations and the area calculation. The original poster expresses uncertainty about whether to multiply the area found by a factor related to the symmetry of the ellipse.

spectravoid
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Find the area under the parametric curve (clearly a circle):
x=5sin(t)
y=2cos(t)
domain of t: [0,2pi]



h(t)=x
g(t)=y

integral from alpha to beta (lower to upper) of g(t)h'(t)dt



integral from 0 to pi/2 of 10cos(t)^2
10(x/2+sin2x/4) evaluated from 0 to pi/2

solution: 2.5pi

however, I'm not sure if that value gets multiplied by either 2 or 4 or is fine as is
 
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Your parametric equations don't describe a circle: they describe an ellipse.

What's your reasoning for integrating g(t)h'(t)? That would be y*dx/dt, and I don't see how that ties into your ellipse.
 
sorry i meant an ellipse

apparently that's the formula

integral of alpha to beta of g(t)h'(t)dt
 
The problem says 0 to 2*pi, which is all the way around the ellipse. You did 0 to pi/2 which is only a quarter of the way around the ellipse. Since the other three quarters look the same as the one quarter that you did, what would be the answer for the entire ellipse?
 
By the way you can look up the formula for the area of an ellipse and check your answer. They should be the same.
 
I was almost tempted to make the mistake of interpreting t as the polar angle and using the area formula for polar coordinates. :blushing:
 
Chrisas said:
The problem says 0 to 2*pi, which is all the way around the ellipse. You did 0 to pi/2 which is only a quarter of the way around the ellipse. Since the other three quarters look the same as the one quarter that you did, what would be the answer for the entire ellipse?

Thanks what I'm confused with

i assumed you multiply that solution by 4

however, i looked up the formula for the area of an oval

It's the length of the major axis/2 x length of the minor axis/2 x pi

therefore the answer would be 2.5pi

that is equal to when i evaluated it from 0 to pi/2.
 
The length of the major axis of this particular ellipse is 10 and it's minor axis length is 4.
 
  • #10
Random Variable said:
The length of the major axis of this particular ellipse is 10 and it's minor axis length is 4.


Well that explains all of my confusion

thanks

i accidentally only looked at half of the minor axis and divided by 2
 

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