Find the length of the given curve. x=3y^(4/3)–(3/23)y^(2/3) –64<=y<=343

Find the length of the given curve. x=3y^(4/3)–(3/23)y^(2/3) –64<=y<=343 (<= is less than or equal to)

Assuming that x = 3y^(4/3) - (3/32)y^(2/3):
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Since the curve is symmetric about the x-axis, it suffices to the add the following arc lengths:
(i) y in [0, 64], and (ii) y in [0, 343].
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dx/dy = 4y^(1/3) - (1/16)y^(-1/3)

So, √(1 + (dx/dy)^2)
= √(1 + (16y^(2/3) - 1/2 + (1/256)y^(-2/3)))
= √(16y^(2/3) + 1/2 + (1/256)y^(-2/3))
= √(4y^(1/3) + (1/16)y^(-1/3))^2
= 4y^(1/3) + (1/16)y^(-1/3), since we are assuming y > 0.

Hence, the arc length equals
∫(y = 0 to 64) (4y^(1/3) + (1/16)y^(-1/3)) dy + ∫(y = 0 to 343) (4y^(1/3) + (1/16)y^(-1/3)) dy
= (1/16) [∫(y = 0 to 64) (64y^(2/3) + 1) y^(-1/3) dy + ∫(y = 0 to 343) (64y^(2/3) + 1) y^(-1/3) dy]
= (1/16) [∫(u = 0 to 16) (64u + 1) * (3/2)du + ∫(u = 0 to 49) (64u + 1) * (3/2)du], letting u = y^(2/3)
= (3/32) [∫(u = 0 to 16) (64u + 1) du + ∫(u = 0 to 49) (64u + 1) du]
= (3/32) [(32u^2 + u) {for u = 0 to 16} + (32u^2 + u) {for u = 0 to 49}]
= 255267/32.

I hope this helps!
I hope this helps!

I feel like there's something off about how you're stating this.

To find a length of a curve, you'd first need to uniquely specify the curve. I'm not sure what you're specifying. It looks kind of like a region, or it looks like some rules about the part of the curve that's of interest got garbled in trying to type in ASCII characters.

=96