Parametric Equations--Area

Question

Find the area bounded by the curve and the line y=65/8:

x = t- 1/t

y= t+ 1/t

I got the area to be 4095/32, but it's wrong. I would appreciate if you could solve it step-by-step, so that I can see where I made a mistake. Thanks!

Answer 1

It is worth exploring the curve on Desmos. Type (t-1/t,t+1/t) and the range is 1/8 to 8. That way you get a sense of the curve. The range can be larger, but those t values are the ones that have y = 65/8

The integral for the area is int(dA) = int(hdx) = int(65/8 -y)dx/dt)dt = int(65/8- t -1/t)(1+1/t²)dt =

You need the limits which you can find solving 65/8 = t + 1/t or 65/8 t = t²+1 If you solve the quadratic in t, you get t = 8 or 1/8

The integral for the area is 55.667 when solved numerically or term by term.

Int (65/8 - 65/8*1/t² -t - 2/t - 1/t³)dt

65/8 *t +65/(8t) - t²/2 - 2ln(t) + 1/(2t²) from 1/8 to 8

Messy problem... the yx box - int(ydx) instead. Have to be careful about what area you are calculating.

Answer 2

corrected algebra

isn't the curve for y>0

y=√(4+x²), the top half of hyperbola

the intercepts being +/- 7.785 on x-axis

where it meets y=65/8

isn't area = [2*(65/8)*7.785] - 2*∫√(4+x²)dx from x=0 to x=7.785

for the integral;. can use x=2sinh(θ) substitution and θ from 0 to 2.08

calculates to 55.59 square units