Area between two curves problem

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SUMMARY

The discussion focuses on calculating the area A between the curves y = x^(1/3) and y = j in the first quadrant. The area A is defined as a function of j, with the integral A(j) = ∫(j - x^(1/3))dx from 0 to j^3. For j = 2, the area A(2) is computed to be 4. Additionally, the discussion raises a question about determining the rate of change of A with respect to k when k = 2, emphasizing the need for differentiation techniques.

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Homework Statement



(a) Find the area A, as a function of j, of the region in the 1st quadrant enclosed by the y-axis and the graphs of y = x ^ (1/3) and y = j for j > 0.

(b) What is the value of A when k = 2

(c) If the line y = k is moving upward at the rate of 1/2 units per second, at what rate is A changing when k = 2?



2. The attempt at a solution

Ok, so I doubt this is right as I was really, confused, but this is what I did...

(a)
x^(1/3) = j
x = j^3

A(j) = integral of (j - x^(1/3))dx from 0 to j^3

(b) A(2) = integral of (2 - x^(1/3))dx from 0 to 2^3
this equals = 4

(c) I have no idea..
I took the derivative of A(x) and got zero...

Please help. :)
 
Physics news on Phys.org
You computed corretcly the Area, which is a function of ? (name it)

If j changes, and you have a certain function of j, how do you compute the change of f(j) ?
 

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