Question 1183479
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You missed it by "THAT" much. Your part A is correct.  But I suspect you used:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \int_1^2\,\sqrt{1\,+\,\(\frac{3y^2}{y^3\,+\,10}\)^2}\,dy]


For your computation of part B.  But remember, in this case, *[tex \Large x] is a function of *[tex \Large y], but *[tex \Large \[1,2\]] is the interval on the *[tex \Large x]-axis.  Hence your limits of integration should be *[tex \Large y(1)] to *[tex \Large y(2)].  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \int_{-\sqrt[3]{10-e}}^{-\sqrt[3]{10-e^2}}\,\sqrt{1\,+\,\(\frac{3y^2}{y^3\,+\,10}\)^2}\,dy]


You might be off a bit in your calculation of the final approximation because you will be forced to use approximations for the integration limits, but you should be a lot closer than the answer you did get for part B.  Wolfram Alpha gives the same answer out to the 15 digits you specified in Part A.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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