The question I am trying to work on is #27 in the image I uploaded. This problem

Question

The question I am trying to work on is #27 in the image I uploaded. This problem has already been solved on cramster (James Stewart - Essential Calculus ((red book)); 9.2).

I know the formula to find the area of this curve is the integral of y*dx from a to b. This being said, I don't understand why the integral of (y-1)dx was evaluated instead. The rest of my work looks the same. Even my limits are the same. I haveeverything right except for the y-1 part. Why is it y-1 and not simply y? Please help!!!

Explanation / Answer

Area = Sy.dx

x = a cos

=> dx = -a sin d

Area = S bsin * (-asin) d from 0< < 2

Thus Area = -ab S sin2 d = -ab S ( 1-cos2) /2 d = -ab (2 - 0) + (ab * ( Sin2 - sin0 )/2) = 2ab

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.