Question

Find the centroid of the region in the first quadrant bounded by the given curves (check my work, where I messed up ? ).

Answer 1

\[y = x^7, x = y^7\]

Answer 2

\[A = \int\limits_{0}^{1}(x^{1/7}-x^7)dx = [\frac{ 7 }{ 8 }x^{8/7}-\frac{ 1 }{ 8 }x^8]_{0}^{1}=\frac{ 7 }{ 8 }-\frac{ 1 }{ 8 }=\frac{ 3 }{ 4 }\]\[X = \frac{ 1 }{ A }\int\limits_{a}^{b}x(f(x)-g(x))dx\]\[X = \frac{ 1 }{ \frac{ 3 }{ 4 } }\int\limits_{0}^{1}x(x^{1/7}-x^7)dx=\frac{ 4 }{ 3 }\int\limits_{0}^{1}(x^{8/7}-x^8)dx=\frac{ 4 }{ 3 }[\frac{ 7 }{ 15 }x^{15/7}-\frac{ 1 }{ 9 }x^9]_{0}^{1}\]\[=\frac{ 4 }{ 3 }[\frac{ 7 }{ 15 }-\frac{ 1 }{ 19 }] = \frac{ 4 }{ 3 }[\frac{ 21 }{ 45 }-\frac{ 5 }{ 45 }]=\frac{ 4 }{ 3 }[\frac{ 16 }{ 45 }]=\frac{ 64 }{ 135 }\]

Answer 3

\[Y = \frac{ 1 }{ A }\int\limits_{0}^{1}[\frac{ 1 }{ 2 }(f(x)^2-g(x)^2)]dx\]\[\frac{ 4 }{ 3 }*\frac{ 1 }{ 2 }\int\limits_{0}^{1}[(x^{1/7} )^2-(x^7)^2]dx = \frac{ 2 }{3 }\int\limits_{0}^{1}(x^{1/49}-x^{49})dx=\frac{ 2 }{ 3 }\int\limits_{0}^{1}[\frac{ 49 }{ 50 }x^{50/49}-\frac{ 1 }{ 50 }x^{50}]_{0}^{1}\]\[=\frac{ 2 }{ 3 }[\frac{ 49 }{ 50 }-\frac{ 1 }{ 50 }]=\frac{ 2 }{ 3 }[\frac{ 48 }{ 50 }]=\frac{ 48 }{ 75 }\]

\[( X, Y) = (\frac{ 64 }{ 135 },\frac{ 48 }{ 75 })\]

My answer is wrong, where did I messed up?

Answer 4

Just looking it over now :)

Answer 5

Ahh there it is

Hint* What is \(\large (x^2)^3\) ? Is it \(\large x^8\) or is it \(\large x^6\) ?

Answer 6

Once you answer that question...take another look at your 'Y' calculation