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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]

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

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