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Find the length of the curve:

[tex]x = 3y^{4/3}-((3/32)y^{2/3})[/tex]

And -343 <= y <= 125.

I used the formula:

[tex]\int_{a}^{b}\ (1 + (x')^{2})^{1/2}dx[/tex]

[tex]\int_{-343}^{125}\ (1+(4y^{1/3}-(1/16)y^{-1/2})^{2})^{1/2}dx[/tex]

But how do you find this integral? Is this even the right integral to use? Thanks.