Question

find the centroid of the region bounded by the given curves y=x^3, x+y=10 and y=0

Answer 1

Solve using limits, 0 to 2 and 2 to 10.

ex: M = From 0 to 2 (x^3)dx + From 2 to 10 (10-x)dx

and so on...
That was the biggest problem for me is figuring out the limits, but if you need more help here is an example solution

Answer 2

The centroid of the region will be "$(\frac{584}{135} , \frac{496}{189} )$".

Given curves:

• $y=x^3$
• $x+y = 10$
• $y =0$

Now,

→ $A = \int\limits^2_0 {x^3} \ dx + \int\limits^2_{10} {10-x} \ dx$

      $= [\frac{1}{4} x^4]^2_0+[10x - \frac{1}{2}x^2 ]^{10}_2$

      $= 4+(100-20)-(50-2)$

      $= 36$

then,

→ $\bar{x} = \frac{1}{A} \int\limits^8_0 \frac{1}{2} ((10-y)^2-y^{\frac{2}{3} })dy$

      $= \frac{1}{2A} \int\limits^8_0 ( 100-20y+y^2-y^{\frac{2}{3} })dy$

      $= \frac{1}{2A} [100y-10y^2+\frac{1}{3}y^3 -\frac{3}{5}y^{\frac{5}{3} } ]^8_0$

      $= \frac{584}{135}$

→ $\bar{y} = \frac{1}{A} \int\limits^8_0 y (10-y-y^{\frac{1}{3} })dy$

     $= \frac{1}{A} \int\limits^8_0 (1oy-y^2-y^{\frac{4}{3} })dy$

     $= \frac{1}{A} [5y^2-\frac{1}{3} y^3 - \frac{3}{7}y^{\frac{7}{3} } ]^8_0$

     $= \frac{496}{189}$

Thus the above answer is appropriate.

Learn more about centroid here:

https://brainly.com/question/5872501