# Thread: Three Integral Calculus Problems

1. ## Three Integral Calculus Problems

Hi, I am currently solving some practice problems and would like some help on these three questions with solutions:

1.) A cross section of a trough is parabolic segment 8 feet wide and 4 feet deep. If the trough is filled with a liquid weighing 45 pounds per cubic foot, find the total force on one end. (ANS: 1536 lb)

2.) Find the volume of the solid generated by revolving about y=2 the region bounded by y2=8x, x=8, and y=2.

3.) Considering two curves: x2+y2=48 and x2+8y=0. Determine the length of the line AB. A and B are points of intersection and also determine the area bounded.

2. ## Re: Three Integral Calculus Problems Originally Posted by mkaraline Hi, I am currently solving some practice problems and would like some help on these three questions with solutions:

1.) A cross section of a trough is parabolic segment 8 feet wide and 4 feet deep. If the trough is filled with a liquid weighing 45 pounds per cubic foot, find the total force on one end. (ANS: 1536 lb)

2.) Find the volume of the solid generated by revolving about y=2 the region bounded by y2=8x, x=8, and y=2.

3.) Considering two curves: x2+y2=48 and x2+8y=0. Determine the length of the line AB. A and B are points of intersection and also determine the area bounded.
Please let us know what you've been able to do on these. That will help us to help you better.

-Dan

3. ## Re: Three Integral Calculus Problems

Hi, so far I have none and the questionnaire only had answers on them.

4. ## Re: Three Integral Calculus Problems

Have you even tried? Every Calculus book I have seen discussed the fact that force on a surface is, for constant pressure, the product of the pressure times the area of the surface- and if the pressure is a function of height, then the total force is the integral of the pressure function over the area.
The force (weight) of a column of water is the density of water times the volume of the column: density time area of the base times height. Dividing the weight by the base area gives pressure as density times height. Finally water exerts the same pressure in all directions (that is pretty much the definition of "fluid") so a column of water exerts pressure against a point on a side wall equal to the density of water times the depth of that point below the water.

That is what you need to solve the first problem.

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