No problem, glad I could help!

[MathewsMD]

Find the values of c such that the area of the region bounded by the parabolas y = x² - c² and y = c² - x² is 576.

Attempt:

576 = _-c^c∫-x² + c² - (x² - c²) dx

576 = 2_-c^c∫c² - x² dx

576 = c²x -(1/3)(x³) l₀^c *I know by symmetry that the area of 0 → c is half the area of -c → c

576 = c³ - (1/3)c³

576 = (2/3)(c³)

c ~ 9.52

This is the incorrect answer for c. I know there are other methods to solve this problem, but I am trying to answer this question using this strategy. Can anyone please point out the error in my work?
Thank you!

 

[Chestermiller]

Find the values of c such that the area of the region bounded by the parabolas y = x² - c² and y = c² - x² is 576.

Attempt:

576 = _-c^c∫-x² + c² - (x² - c²) dx

576 = 2_-c^c∫c² - x² dx

576 = c²x -(1/3)(x³) l₀^c *I know by symmetry that the area of 0 → c is half the area of -c → c

576 = c³ - (1/3)c³
You did the integration incorrectly. First you lost your original factor of 2. Then, you lost another factor of 2 when you forgot to substitute the lower integration limit.

576 = (2/3)(c³)

c ~ 9.52

This is the incorrect answer for c. I know there are other methods to solve this problem, but I am trying to answer this question using this strategy. Can anyone please point out the error in my work?
Thank you!

You did the integration incorrectly. First you lost your original factor of 2. Then, you lost another factor of 2 when you forgot to substitute the lower integration limit. Just redo the integration with more care, and you'll get the right answer.

 

[MathewsMD]

You did the integration incorrectly. First you lost your original factor of 2. Then, you lost another factor of 2 when you forgot to substitute the lower integration limit. Just redo the integration with more care, and you'll get the right answer.

Haha okay, I actually canceled the factors out (I must have thought it was in the denominator for some reason). Thank you for finding it!