Question

Find the centroid of the solid generated if the area bounded by
xy=4,x=2,x=4,y=0 revolving about x-axis

Accepted Answer

SOLUTION:

The curves are given as:

xy=4, x=2, x=4, y=0 is revolved around x-axis.

Plotting the curves and finding the intersection point.

[/image?k=d9d0474328205b11858264eff039621f]

xy=4, x=2, x=4, y=0

At x=2

xy=4

y=\frac{4}{x} \y=\frac{4}{2}

y=2

At x=4

xy=4

y=\frac{4}{x} \y=\frac{4}{4}

y=1

So, coordinates are (2,2) and (4,1)

Now, mass is:

M=\iint_{D} p(x, y) \mathrm{d} A

x varies from x=2 to x=4 y varies from y=0 to y=\frac{4}{x}

\M=\iint_{D} p(x, y) \mathrm{d} A \M=\int_{x=2}^{x=4}
\int_{y=0}^{y=\frac{4}{x}} \quad \mathrm{~d} y \mathrm{~d} x
\M=\int_{2}^{4} \int_{0}^{\frac{4}{x}} \mathrm{~d} y \mathrm{~d} x
\M=\int_{2}^{4}[y]_{0}^{\frac{4}{x}} \mathrm{~d} x
\M=\int_{2}^{4}\left[\frac{4}{x}-0ight] \mathrm{d} x
\M=\int_{2}^{4} \frac{4}{x} \mathrm{~d} x \M=4[\ln (x)]_{2}^{4}
\M=4[\ln (4)-\ln (2)] \M=4 \ln \left(\frac{4}{2}ight) \M=4 \ln
(2)

Next step is Calculating x Coordinate

\\bar{x}=\frac{1}{M} \iint_{D} x \cdot p(x, y) \mathrm{d} A
\\bar{x}=\frac{1}{4 \ln (2)} \int_{x=2}^{x=4}
\int_{y=0}^{y=\frac{4}{x}} x \mathrm{~d} y \mathrm{~d} x
\\bar{x}=\frac{1}{4 \ln (2)} \int_{2}^{4} \int_{0}^{\frac{4}{x}} x
\mathrm{~d} y \mathrm{~d} x \\bar{x}=\frac{1}{4 \ln (2)} \int_{2}^{4}
x[y]_{0}^{\frac{4}{x}} \mathrm{~d} x \\bar{x}=\frac{1}{4 \ln (2)}
\int_{2}^{4} x\left[\frac{4}{x}-0ight] \mathrm{d} x
\\bar{x}=\frac{1}{4 \ln (2)} \int_{2}^{4} 4 \mathrm{~d} x
\\bar{x}=\frac{4}{4 \ln (2)}[x]_{2}^{4} \\bar{x}=\frac{1}{\ln
(2)}[4-2] \\bar{x}=\frac{2}{\ln (2)} \Putting\ \ln (2)=0.69315.,\
we\ get \\bar{x}=\frac{2}{\ln (2)} \\bar{x}=\frac{2}{0.69315}
\\bar{x}=2.88538 \approx 2.885

Now, Calculating y Coordinate

\bar{x}=\frac{1}{M} \iint_{D} y \cdot p(x, y) \mathrm{d} A
\\bar{y}=\frac{1}{4 \ln (2)} \int_{x=2}^{x=4}
\int_{y=0}^{y=\frac{4}{x}} y \mathrm{~d} y \mathrm{~d} x
\\bar{y}=\frac{1}{4 \ln (2)} \int_{2}^{4} \int_{0}^{\frac{4}{x}} y
\mathrm{~d} y \mathrm{~d} x \\bar{y}=\frac{1}{4 \ln (2)}
\int_{2}^{4}\left[\frac{y^{2}}{2}ight]_{0}^{\frac{4}{x}} \mathrm{~d}
x \\bar{y}=\frac{1}{4 \ln (2)}
\int_{2}^{4}\left[\frac{\left(\frac{4}{x}ight)^{2}-(0)^{2}}{2}ight]
\mathrm{d} x \\bar{y}=\frac{1}{4 \ln (2)}
\int_{2}^{4}\left[\frac{16}{2 x^{2}}ight] \mathrm{d} x
\\bar{y}=\frac{1}{4 \ln (2)} \int_{2}^{4} \frac{8}{x^{2}} \mathrm{~d}
x \\bar{y}=\frac{8}{4 \ln (2)} \int_{2}^{4} \frac{1}{x^{2}}
\mathrm{~d} x \\bar{y}=\frac{2}{\ln
(2)}\left[\frac{x^{-2+1}}{-2+1}ight]_{2}^{4} \\bar{y}=\frac{2}{\ln
(2)}\left[\frac{x^{-1}}{-1}ight]_{2}^{4} \\bar{y}=\frac{2}{\ln
(2)}\left[-\frac{1}{x}ight]_{2}^{4} \\bar{y}=\frac{2}{\ln
(2)}\left[\left(-\frac{1}{4}ight)-\left(-\frac{1}{2}ight)ight]
\\bar{y}=\frac{2}{\ln (2)}\left[-\frac{1}{4}+\frac{1}{2}ight]
\\bar{y}=\frac{2}{\ln (2)}\left[\frac{-1+2}{4}ight]
\\bar{y}=\frac{2}{\ln (2)}\left(\frac{1}{4}ight)
\\bar{y}=\frac{1}{2 \ln (2)}

Putting \ln (2)=0.69315 we get,

\bar{y}=\frac{1}{2(0.69315)} \\bar{y}=\frac{1}{1.3863}
\\bar{y}=0.72134 \approx 0.721

So, the Centroid is: (\bar{x}, \bar{y})=(2.885,0.721)

Question #229267

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