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## Homework Statement

Find the centroid of the region bounded by the given curves.

[tex]y = x^3[/tex][tex]x + y = 10[/tex][tex]y = 0[/tex]

## Homework Equations

[tex]\bar x = \frac{1}{A}\int^b_a xf(x) dx[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(x))^2 dx[/tex]

Where A is the area of the region containing the centroid.

## The Attempt at a Solution

First thing I did was find the intersection of the curves [tex]x^3 = 10-x \Rightarrow x = 2 \Rightarrow P(2,8)[/tex]

Then I put things in terms of y to integrate to find the area A of the region.[tex]y = x^3 \Rightarrow x = \sqrt[3] y[/tex][tex]x + y = 10 \Rightarrow x = 10 - y[/tex][tex]A = \int^8_0 ((10-y)-(\sqrt[3] y)) dy = 36[/tex]

So [tex]\bar x = \frac{1}{A}\int^b_a yf(y) dy = \frac{1}{36}\int^8_0 y((10-y)-(\sqrt[3] y))dy = \frac{496}{189} \approx 2.63[/tex][tex]\bar y = \frac{1}{A}\int^b_a \frac{1}{2}(f(y))^2 dy = \frac{1}{36}\int^8_0 \frac{1}{2}((10-y)^2-(\sqrt[3] y)^2)dy = \frac{584}{135} \approx 4.33[/tex]

So the centroid is at [tex]\left(\frac{496}{189},\frac{584}{135}\right)[/tex]

What I'm getting here isn't correct. Thanks for commenting!