Question

Sketch the region bounded by the given curve and visually estimates the centroid location and the exact coordinates of the centroid \((\bar x,\bar y)\).

Short answer

The location \((\bar x,\bar y)\) of the region bounded by the given curve is \((2.3,0.8)\). The exact coordinates of the centroid \((\bar x,\bar y)\) is \(\left( {\frac{{12}}{5},\frac{3}{4}} \right)\).

Step-by-step solution

Plot a graph for the equation \(y = \sqrt x \)and x=4

Calculate value of y

Substitute 0 for \(x\)

\(y = \sqrt 0 \)

y=0

Hence the co-ordinate of (x, y) is \((0,0)\).

Calculate value of y

Substitute 1 for \(x\)

\(\begin{aligned}{}y = \sqrt 1 \\y = \pm 1\end{aligned}\)

Hence, the co-ordinate of \((x,y)\) is \((1, \pm 1)\).

Calculate value of \(y\)

Substitute 2 for \(x\)

\(y = \sqrt 2\)

\(y = \pm 1.414\)

Hence the co-ordinate of (x, y) is \((2, \pm 1.414)\).

Calculate value of \(y\)

Substitute 3 for \(x\)

\(y = \sqrt 3\)

\(y = \pm 1.732\)

Hence, the co-ordinate of \((x,y)\) is \((3, \pm 1.732)\).

Calculate value of \(y\)

Substitute 4 for \(x\)

\(\begin{aligned}{}y = \sqrt 4 \\y = \pm 2\end{aligned}\)

Hence the co-ordinate of (x, y) is \((4, \pm 2)\).

Find the area of the region.

\( A = \int_0^4 {\sqrt x } dx\)

\( A = \frac{2}{3}\left[ {{x^{3/2}}} \right]_0^4\)

\( A = \frac{2}{3}\left( {{4^{3/2}} - 0} \right)\)

\( A = \frac{2}{3} \cdot 8\)

\( = \frac{{16}}{3}\)

Find the \(x\)-coordinate of the centroid.

\( \bar x = \frac{1}{A}\int_a^b x f(x)dx\)

\( \bar x = \frac{1}{{\frac{{16}}{3}}}\int_0^4 x (\sqrt x )dx\)

\( \bar x = \frac{3}{{16}}\int_0^4 {{x^{3/2}}} dx\)

\( \bar x = \frac{3}{{16}} \cdot \frac{2}{5}\left[ {{x^{5/2}}} \right]_0^4\)

\( \bar x = \frac{3}{{16}} \cdot \frac{2}{5}\left[ {{4^{5/2}} - 0} \right]\)

\( \bar x = \frac{3}{{16}} \cdot \frac{2}{5}(32)\)

\( \bar x = \frac{3}{1} \cdot \frac{2}{5}(2)\)

\( \bar x = \frac{{12}}{5}\)

Find the \(y\)-coordinate of the centroid.

\( \bar y = \frac{1}{A}\int_a^b {\frac{1}{2}} {[f(x)]^2}dx\)

\( \bar y = \frac{3}{{16}} \cdot \frac{1}{2}\int_0^4 {{{(\sqrt x )}^2}} dx\)

\( \bar y = \frac{3}{{32}}\int_0^4 x dx\)

\( \bar y = \frac{3}{{32}} \cdot \frac{1}{2}\left[ {{x^2}} \right]_0^4\)

\( \bar y = \frac{3}{{64}}[16 - 0]\)

\( \bar y = \frac{3}{4}\)