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

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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Mark44
Mentor
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. 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.

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