Question

the area enclosed between the x-axis, the curve z=x(3-x)and the coordinates x=1 and x=4 is rotated through 2pi radians about the x- axis.calculate : a)the rotating area and its centroid ?? please help before the deadline :s ?

Answer 1

so it is really a hard question i thought i am the only one who doesn't know how to solve it

Answer 2

this problem requires integration, the given function is a parabola open downward...
we can simply find the rotating area by integrating the given function and evaluate it at given boundaries, from x=1 to x=4....

Answer 3

\[\int\limits_{1}^{4} (3x-x ^{2}) dx=\int\limits_{1}^{4}3xdx-\int\limits_{1}^{4}x ^{2}dx\]

Answer 4

\[Area=\frac{ 3 }{ 2 }\left[ x ^{2} \right]_{1}^{4}-\frac{ 1 }{ 3 }\left[ x ^{3} \right]_{1}^{4}\]

Answer 5

\[Area=\frac{ 3 }{ 2 }\left[ (4)^{2}-(1)^{2} \right]-\frac{ 1 }{ 3 }\left[ (4)^{3}-(1)^{3} \right]\]
\[Area=\frac{ 3 }{ 2 }(16)-\frac{ 1 }{ 3 }(64)=24-21.333=2.667\]

Answer 6

The area is 2.667 square units....

Answer 7

thank u so much

Answer 8

but what its centroid is ?

Answer 9

this is the question

Answer 10

in finding the centroids... we can make use of this site as a reference...
http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

Answer 11

the Centroid coordinates:
\[x = \frac{ M _{x} }{ M }=\frac{ 1 }{ Area } \int\limits_{a}^{b} x[f(x)-g(x)]dx\]
\[y=\frac{ M _{y} }{ M }=\frac{ 1 }{ Area }\int\limits_{a}^{b}\frac{ 1 }{ 2 }([f(x)]^{2}-[g(x)]^{2})dx\]

Answer 12

\[f(x) = x(3-x) =3x-x ^{2}\]
\[g(x)=z=0\]

Answer 13

substitute f(x) and g(x) to the above formula, again evaluate at given limits... we can determine the centroid of our rotating area....