Solution.
S = 1 4 π r 2 − 1 2 • 5 • 5 = 25 4 π − 25 2 = 7. S=\frac{1}{4}πr^2-\frac{1}{2}•5•5=\newline
\frac{25}{4}π-\frac{25}{2}=7. S = 4 1 π r 2 − 2 1 •5•5 = 4 25 π − 2 25 = 7.
x C = 1 S ∫ 0 5 ( x ( 25 − x 2 − ( 5 − x ) ) d x = = 1 7 ∫ 0 5 ( x 25 − x 2 − 5 x + x 2 ) ) d x = = 1 7 • 125 6 = 125 42 . x_C=\frac{1}{S}\int_0^5 (x(\sqrt{25-x^2}-(5-x))dx=\newline
=\frac{1}{7}\int_0^5 (x\sqrt{25-x^2}-5x+x^2))dx=\newline
=\frac{1}{7}•\frac{125}{6}=\frac{125}{42}. x C = S 1 ∫ 0 5 ( x ( 25 − x 2 − ( 5 − x )) d x = = 7 1 ∫ 0 5 ( x 25 − x 2 − 5 x + x 2 )) d x = = 7 1 • 6 125 = 42 125 .
y C = 1 2 S ∫ 0 5 ( 25 − x 2 − ( 5 − x ) 2 ) d x = 1 14 ∫ 0 5 ( − 2 x 2 + 10 x ) d x = 1 14 ( − 2 3 x 3 + 5 x 2 ) ∣ 0 5 = 1 14 • 125 3 = 125 42 . y_C=\frac{1}{2S}\int_0^5(25-x^2-(5-x)^2)dx=\newline
\frac{1}{14}\int_0^5(-2x^2+10x)dx=\newline
\frac{1}{14}(-\frac{2}{3}x^3+5x^2)|_0^5=\newline
\frac{1}{14}•\frac{125}{3}=\frac{125}{42}. y C = 2 S 1 ∫ 0 5 ( 25 − x 2 − ( 5 − x ) 2 ) d x = 14 1 ∫ 0 5 ( − 2 x 2 + 10 x ) d x = 14 1 ( − 3 2 x 3 + 5 x 2 ) ∣ 0 5 = 14 1 • 3 125 = 42 125 .
So, centroid ( x C , y C ) = ( 125 42 , 125 42 ) . (x_C,y_C)=(\frac{125}{42},\frac{125}{42}). ( x C , y C ) = ( 42 125 , 42 125 ) .
Answer. ( 125 42 , 125 42 ) . (\frac{125}{42},\frac{125}{42}). ( 42 125 , 42 125 ) .
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