Answer:
The centroid is the point ([tex]\frac{584}{135},\frac{496}{189}[/tex])=(4.32,2.62)
Step-by-step explanation:
We graph the functions given in the attached picture. As you can see the region R is surrounded by the functions:
y=x³
y=10-x
y=0
This region can be divided by 2, where the first is delimited by y=x³ (0≤x≤2) and the second is y=10-x (2≤x≤10).
The centroid of region is given by:
[tex]x_c=\frac{\int\limits_R {x} \, dydx }{\int\limits_R \, dydx}[/tex]
[tex]y_c=\frac{\int\limits_R {y} \, dydx }{\int\limits_R \, dydx}[/tex]
First, we have to find the value of [tex]\int\limits_R \, dydx[/tex]
[tex]\int\limits_R \, dydx=\int\limits^2_0 {\int\limits^{x^{3}}_0} \, dydx +\int\limits^{10}_2 {\int\limits^{10-x}_0} \, dydx=\int\limits^2_0 {y|^{x^{3}}_{0}} \, dx +\int\limits^{10}_2 {y|^{10-x}_{0}} \, dx[/tex]
=[tex]\int\limits^2_0 {x^3} \, dx +\int\limits^{10}_2 {(10-x)} \, dx=\frac{x^4}{4} |^{2}_{0}+(10x-\frac{x^2}{2})|^{10}_{2}[/tex]
=(16/4-0)+(100-50-20+2)=4+32=36
Now we find [tex]\int\limits_R {x} \, dydx[/tex]:
[tex]\int\limits_R {x} \, dydx=\int\limits^2_0 {\int\limits^{x^{3}}_0} {x}\, dydx +\int\limits^{10}_2 {\int\limits^{10-x}_0}{x} \, dydx=\int\limits^2_0 {x(y)|^{x^3}_0}\, dx +\int\limits^{10}_2 {x(y)|^{10-x}_0} \, dx[/tex]
=[tex]\int\limits^2_0 {x^4}\, dx +\int\limits^{10}_2 {(10x-x^2)} \, dx=\frac{x^5}{5} |^{2}_{0} +(5x^2-\frac{x^3}{3} )|^{10}_2[/tex]
=[tex](\frac{32}{5}-0)+(500-\frac{1000}{3}-20+\frac{8}{3})=\frac{2336}{15}[/tex]
Then, we find [tex]\int\limits_R {y} \, dydx[/tex]:
[tex]\int\limits_R {x} \, dydx=\int\limits^2_0 {\int\limits^{x^{3}}_0} {y}\, dydx +\int\limits^{10}_2 {\int\limits^{10-x}_0}{y} \, dydx=\int\limits^2_0 {(\frac{y^2}{2})|^{x^3}_0}\, dx +\int\limits^{10}_2 {(\frac{y^2}{2})|^{10-x}_0} \, dx[/tex]
=[tex]\int\limits^2_0 {\frac{x^6}{6}}\, dx +\int\limits^{10}_2 {\frac{(10-x)^2}{2}} \, dx=\int\limits^2_0 {\frac{x^6}{2}}\, dx +\int\limits^{10}_2 {(50-10x+\frac{x^2}{2})} \, dx\\=\frac{x^7}{14} |^{2}_{0} +(50x-5x^2+\frac{x^3}{6} )|^{10}_2[/tex]
=[tex]\frac{x^7}{14} |^{2}_{0} +(50x-5x^2+\frac{x^3}{6} )|^{10}_2=(\frac{64}{7}-0)+(500-500+\frac{500}{3})-(100-20+\frac{4}{3})=\frac{1984}{21}[/tex]
Hence,
[tex]x_c=\frac{\frac{2336}{15}}{36}=\frac{584}{135}[/tex]
[tex]y_c=\frac{\frac{1984}{21}}{36}=\frac{496}{189}[/tex]
The centroid is the point ([tex]\frac{584}{135},\frac{496}{189}[/tex])=(4.32,2.62)
