Thread: Calc 1 area of plane curve

1. Calc 1 area of plane curve

Consider the parametric equation:

x = 3(cos(theta) + (theta)sin(theta))
y = 3(sin(theta) - (theta)cos(theta))

what is the length of the curve for theta = 0 to theta = 3/2(pi)?

I put the derivative of x and y in a length formula to get: I'll put "t" for theta

int_0^3/2pi sqrt[(3(tcost))^2 + (3(-tsint))^2]

then i distributed the squares to get:

int_0^3/2pi sqrt[9t^2(cos^2t+sin^2t)]

then i took out the 9t^2 to get:

3 int_0^3/2pi sqrt[(cos^2t + sin^2t)]

and integrated...

then i took out the 9t^2 to get:

3 int_0^3/2pi sqrt[(sin^2t - cos^2t)]

I put in t = 3/2pi to get: 29.09768846
my online submission was declared incorrect. Can anybody help me see what I did wrong?

Thanks!

2. Everything is correct except what I put in red color

I put the derivative of x and y in a length formula to get: I'll put "t" for theta

int_0^3/2pi sqrt[(3(tcost))^2 + (3(+tsint))^2]

then i distributed the squares to get:

int_0^3/2pi sqrt[9t^2(cos^2t+sin^2t)]

then i took out the 9t^2 to get:

3 int_0^3/2pi t sqrt[(cos^2t + sin^2t)]

and integrated...

then i took out the 9t^2 to get:

3 int_0^3/2pi t sqrt[(sin^2t + cos^2t)]

I put in t = 3/2pi to get: 29.09768846

Remember that sin^2t + cos^2t = 1
Therefore you get
3 int_0^3/2pi t dt = 3/2 (3pi/2)² = 27pi²/8 = 33.31

3. Thank you!

One quick question. If the integral of sinx is -cosx, why is the integral of sin^2x cos^2x and not -cos^2x?

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