# MATH 141 Study Guide - Fall 2018, Comprehensive Term Test Notes - Gravity, Integral, Trigonometric Functions

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Published on 5 Nov 2018
School
Department
Course
University of Maryland
MATH 141
Calculus II
Term Test
Fall 2018
Prof. Patrick Brosnan
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MATH 141 Lecture 5
Tanks
Suppose we have a tank and we want to pump water out of it so that it ends up at a height L
Say the tank is given by region:
 
     
  
  
     
Work required to lift slice of water from z=z0 to z=L is given by:
   
So if a distance exists between a and b, then we get the integral expression:
   
Problem:
A hemispherical tank is filled with water and has a radius of 10 ft. Find work required to pump
the water 6 ft above the tank.
Solution:
The Cross section at z=z0 is a disk
 
 

 
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So,  
 
So, we put the values in the earlier formulae:
  
  

On solving, we get Work Done W = 406250 ft.lbs (ANS)
Kinetic Energy = (1/2) x mass x (velocity)2 = (1/2) mv2
So, If you do work on an object in free motion, it acquires a kinetic energy equivalent to the
work done. This is the Law of Conservation of Energy. Look up the text for its derivation and
how to use this in relation to Newton’s universal Law of Gravitation.
Moments and Center of Gravity
Suppose we have a bunch of masses m1,m2…….mN at various points.
(x1,y1)….(xN,yN) in R2
We want to find a point in the plane about which they will balance.
Definition: The moment of inertia of the above collected points about Y-axis is:
My = m1x1 + m2x2 + ……… + mNxN
Similarly, Moment of inertia about X axis is : Mx = m1y1 + m2y2 + ………… + mNyN
Lets work with m1+m2+m3+ ….. +mN = M
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## Document Summary

Suppose we have a tank and we want to pump water out of it so that it ends up at a height l. Say the tank is given by region: (cid:3409)(cid:1844)(cid:2871) {(cid:4666)(cid:1876),(cid:1877),(cid:1878)(cid:4667)} (cid:1875) (cid:1857)(cid:1870)(cid:1857) (cid:1876),(cid:1877),(cid:1878) (cid:1854)(cid:1857)(cid:1864)(cid:1867)(cid:1866)(cid:1859)(cid:1871) (cid:1872)(cid:1867) (cid:1844) (cid:1845)(cid:1872)(cid:1853)(cid:1870)(cid:1872) (cid:1875)(cid:1872) (cid:3410)(cid:882) (cid:1858)(cid:1867)(cid:1870) (cid:1855)(cid:1870)(cid:1867)(cid:1871)(cid:1871) (cid:1871)(cid:1857)(cid:1855)(cid:1872)(cid:1867)(cid:1866) (cid:1853)(cid:1872) (cid:1852)=(cid:1852)(cid:2868),(cid:1871)(cid:1867) (cid:2868)={(cid:4666)(cid:1876),(cid:1877)(cid:4667)}(cid:1854)(cid:1857)(cid:1864)(cid:1867)(cid:1866)(cid:1859)(cid:1871) (cid:1872)(cid:1867) (cid:1844) (cid:1845)(cid:1867) (cid:1875)(cid:1857) (cid:1853)(cid:1874)(cid:1857) (cid:1870)(cid:1857)(cid:1853) (cid:1867)(cid:1858) (cid:2868)=(cid:4666)(cid:1852)(cid:2868)(cid:4667) So if a distance exists between a and b, then we get the integral expression: A hemispherical tank is filled with water and has a radius of 10 ft. find work required to pump the water 6 ft above the tank. The cross section at z=z0 is a disk (cid:2868)={(cid:4666)(cid:1876),(cid:1877)(cid:4667)}(cid:1875) (cid:1857)(cid:1870)(cid:1857) (cid:1876),(cid:1877) (cid:1854)(cid:1857)(cid:1864)(cid:1867)(cid:1866)(cid:1859)(cid:1871) (cid:1872)(cid:1867) (cid:1844)(cid:2870) So, we put the values in the earlier formulae: So, (cid:4666)(cid:1878)(cid:2868)(cid:4667)=(cid:2024)(cid:4672) (cid:883)(cid:882)(cid:882) (cid:1878)(cid:2868)(cid:2870)(cid:4673)(cid:2870)=(cid:2024)(cid:4666)(cid:883)(cid:882)(cid:882) (cid:1878)(cid:2868)(cid:2870)(cid:4667) (cid:3029) (cid:1849)= (cid:888)(cid:884). (cid:887)(cid:4666)(cid:1878)(cid:4667)(cid:4666) (cid:1878)(cid:4667)(cid:1856)(cid:1878) (cid:3028) (cid:2868) (cid:1849)= (cid:888)(cid:884). (cid:887)(cid:2024)(cid:4666)(cid:883)(cid:882)(cid:882) (cid:1878)(cid:2870)(cid:4667)(cid:4666)(cid:888) (cid:1878)(cid:4667)(cid:1856)(cid:1878) On solving, we get work done w = 406250 ft. lbs (ans)