The portion of the graph between x = 0 and x = 1 is rotated around the y axis to

Question

The portion of the graph

between x = 0 and x = 1 is rotated around the y axis to form a container. The container is filled with water. Use n = 6 subintervals and Simpson's rule to approximate the work required to pump all of the water out over the side of the container. Give your answer in decimal form.
(Distance is measured in meters, the density of water is 1000 kg/m3, and use 9.8 m/s2 for g.)

The portion of the graph y = tan?1 x between x = 0 and x = 1 is rotated around the y axis to form a container. The container is filled with water. Use n = 6 subintervals and Simpson's rule to approximate the work required to pump all of the water out over the side of the container. Give your answer in decimal form. (Distance is measured in meters, the density of water is 1000 kg/m3, and use 9.8 m/s2 for g.)

Explanation / Answer

The radius r of an arbitrary slice I have drawn in red and the distance d through which it must be lifted is drawn in green. We know the amount of work required to lift this slice is the product of the applied force and the distance over which this force is applied, hence:

dW=Fd

Now, the applied force is simply the weight of the slice:

F=mg

and the mass of the slice is the product of its mass density ? and it volume V:

m=?V

and of course the volume of the slice is:

V=?r2dy

where the radius of the slice is:

r=x=tan(y)

Now, putting this together, we find:

dW=Fd=(mg)(?4?y)=(g?V)(?4?y)=g?(?r2dy)(?4?y)=?g?tan2(y)(?4?y)dy

Summing up all the elements of work, we obtain:

W=?g???40tan2(y)(?4?y)dy

Applying Simpson's Rule with n=6 on the integral, where:

f(y)=tan2(y)(?4?y)

and:

yk=?4?06k=?24k with k?{0,1,2,3,4,5,6}

we find:

S6=?4?03(6)[f(y0)+4f(y1)+2f(y2)+4f(y3)+2f(y4)+4f(y5)+f(y6)]

S6=?72[f(y0)+4f(y1)+2f(y2)+4f(y3)+2f(y4)+4f(y5)+f(y6)]

f(y0)=f(?24?0)=f(0)=tan2(0)(?4?0)=0

4f(y1)=4f(?24?1)=4f(?24)=4tan2(?24)(?4??24)?0.045376065047798

2f(y2)=2f(?24?2)=2f(?12)=2tan2(?12)(?4??12)?0.075185401439313

4f(y3)=4f(?24?3)=4f(?8)=4tan2(?8)(?4??8)?0.269506042226324

2f(y4)=2f(?24?4)=2f(?6)=2tan2(?6)(?4??6)?0.174532925199433

4f(y5)=2f(?24?5)=4f(5?24)=4tan2(5?24)(?4?5?24)?0.308290092997036

f(y6)=f(?24?6)=f(?4)=tan2(?4)(?4??4)=0

Adding these together, we find:

S6?0.0380870342601150

For comparison, a numeric integration function returns:

0.038148452745930

Now, using the approximation we obtained, we may state:

W??g??0.0380870342601150

Using the given data:

g=9.8ms2,?=1000kgm3

and observing the integral has units of m4, we have:

W??(9.8ms2)(1000kgm3)?0.0380870342601150 m4?1172.60868088 J

For comparison, the exact answer is:

W=1225?4(16ln(2)??2) J?1174.5 J

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