1. Volume and Work
A container is created by revolving the curve y=x2″ title=”y equals x squared” src=”https://lh4.googleusercontent.com/WJPq_KVsGG8UUgPyELPVR7K5yLLkZfAm8KUf4OYV16GLWt1OZoUFqwyJI2m75-vuzXeHFl71y2XxS78FB-vUZ25ZRIjc7IY72TZlCGcJ3koGht-qzIBIK8yOi_isQs9-2oB86HxT” width=”54″ height=”23″ style=”margin-left: 0px; margin-top: 0px;”> from y=0 to y= 9 about the y-axis.
a. Write an integral that computes V(h), the volume of liquid contained if the container is filled to a height h.
b. How much water does this container hold when it is full?
c. To what height does the water level reach when the volume is half-full?
d. If the container is full of water, how much work does it take to pump all of the water out of the container? Use the symbols ρ” title=”rho” src=”https://lh4.googleusercontent.com/uSmw8P3W7ReX7fXmU4iju7SdXc1vToia52ydLqgXKn8lWRIIDuOQUe90m5zeVZMp9_Xti3Jda1-g0vz6NcYH2eMMw80F_ijjIWWQuwTIMc8VfhWbovlTJzcYhnHHuFXOxLALoPZw” width=”15″ height=”19″ style=”margin-left: 0px; margin-top: 0px;”> and g in your computations to represent density of water and acceleration of gravity. If you’d like a number at the end, you can estimateρ=1000kg/m3″ title=”rho equals 1000 k g slash m cubed” src=”https://lh4.googleusercontent.com/dYMhvk_WoQm38ZfbxpoH_zU4yXp0WFBF_ImZvh-Uh6TrbUcvdPEsXYMsyXpfLrdqYEEn4nkh68EU7iCYFIy8e1HpyZjwfA9htUsxpYM6BlErHHvKTecLbjuF2qB0Z_Ra3DOGeCbq” width=”129″ height=”24″ style=”margin-left: 0px; margin-top: 0px;”> and g=10m/s2″ title=”g equals 10 m slash s squared” src=”https://lh6.googleusercontent.com/MPElEv8zC5RIFTj3plVNsWGqbBG86fyQV0vIBAYg1Hn-jVG-JNkxd_Qob8AO2T6K7qnTxXCEST-Rc3CaG04bp4HPUlAtlnzqIb-TfR-c9YuBhabDdqUaD-GGeSTEDMmcDQFBniqZ” width=”96″ height=”24″ style=”margin-left: 0px; margin-top: 0px;”>
e. Your pump breaks down after pumping out half of the volume of water in the tank. What proportion of the work required to pump all the water out was done? (Hint: use your answer from above, and think about pumping the water out from the top of the tank down to the remaining water level.)
2. Integration by Parts
Let p(x) be an abstract function defined on the interval [a, b] with the following properties:
p(a) = p(b) = 0
p’’(x) exists for each x in [a,b].
a. Use these properties to show that ∫abp(x)p″(x)dx=−∫ab(p′(x))2dx” title=”the integral from a. to b of p of x p double prime of x equals negative the integral from a. to b of open paren p prime of x close paren squared” src=”https://lh6.googleusercontent.com/6i1MWXbQD5suOEkJC1N8AnKOMRH9Mn2IFuBNTPAi_SpcR4nLUIXirQKd-1v1uC4E7T9BeILcD1HQSqJPElG8f-2gObJCj4QxKCC57GqCaZSy2fg7PEFxVavtND-wYu56ApNjBQbf” width=”336″ height=”56″ style=”margin-left: 0px; margin-top: 0px;”>. (Hint, integrate the left side by parts and simplify).
As another side note, it will be useful to use the integration by parts formula if the integrals have limits:
∫abf(x)g′(x)dx+∫abf′(x)g(x)dx=f(x)g(x)|ab” title=”the integral from a. to b of f of x g prime of x plus the integral from a. to b of f prime of x g of x equals f of x g of x divides sub a. to the bth power” src=”https://lh3.googleusercontent.com/f-nKprXTAFJZBUufwVMv-_VOGaL-sQ6QclY4lSLgGRRl35e1QjvbMnN7U3tklpJMKdWK4LX9vNlgirQqOiKSSB0r-XjXLB5y8X-yIuY04vyJLbCRrHKO3t4OXhQeAjSyWNLfznpr” width=”455″ height=”56″ style=”margin-left: 0px; margin-top: 0px;”>. Note that the right side of the equation is evaluated from a to b.
b. In addition to the properties above, suppose p’’ is proportional to p. In symbols. p’’(x) = Kp(x), for some constant K. Show that K must be a negative number.
c. We will apply these results to work on a common integral next week. For now, we just think about functions that satisfy each of the following criteria:
Make up two functions, along with an interval [a, b] for each so that f(a) = f(b) = 0.
Make up two different functions so that f’’(x) = K f(x) for some constant K.
For each of the four functions you wrote down, test each condition.