"8. Do not ask too many questions in one thread. Do not ask more than two questions in a post. "
Alright, so I have my test review in my hand. This semester's instructor doesn't use coursecompass and the demo class we get is useless. He also doesn't answer questions over e-mail. This was fine and all back in the day that there were study groups for calculus up at my school, but there aren't any now. SO! End rant.
Have answers to the first two, it seems!
Spoiler:
3. "The base of a solid is the parabolic region {(x,y)|y^2 <=x<=1} Cross sections perpendicular to the X-axis are squares. Find the volume." The answer is 2.
I've never seen that syntax before and I have no idea what he's asking.
4. "Find the total hydraulic force on a dam in the shape of an equilateral triangle with one vertex pointing down, if the side of the triangle is 100 feet and the water is even with the top. delta = 62 lb/ft^3, P= F/A, P = delta*h" Answer is 7,775,000 LB.
I'll be honest, I do not have enough of a concept of what's going on here. I know we're talking about a triangle with points at x=-100, x=100, and y=-100. I know somewhere in this I use the integral from A to B of w*( strip depth ) * L(y), but I'm sitting here with the verbatim what he worked in class and don't know what's happening.
Any help is appreciated! Thanks guys!
The shape of a solid of revolution can be thought of as a number of small cylinders being stuck together. So its volume can be approximated by summing the volumes of these cylinders. As you make the height of each cylinder smaller and increase the number of cylinders, the approximation converges on the exact volume.
If you are rotating about the axis, the radius of each cylinder at any point is equal to the corresponding value, and call the height of this cylinder , because the height of this cylinder is a small change in .
So the volume of this cylinder is .
The volume of your shape is the sum of all these cylinder volumes, so
and as you make each cylinder's heights smaller and create an infinite sum of cylinders, the sum converges on an integral and the approximation becomes exact.
So , where are the endpoints of .
Do you think you can do Q1 now?
To find surface areas of solids of revolution, divide the region into cylinders and sum the curved surface area of each cylinder.
If you choose two close points and , then each cylinder will have a radius of , where corresponds to .
Call the height of this cylinder , which is the distance between and .
By Pythagoras, , where is the change in and is the change in .
Therefore, the curved surface area of each cylinder is
.
So the surface area of the solid of revolution is
which, as you make small and increase the number of cylinders, this approximation becomes exact and the sum converges on an integral.
So , where are the endpoints of .
Can you do Q2 now?
Got #1. My limits were off. For some reason I went off and took the place where the graph intersects X=1 ( Y=2 ) as my upper limit of integration. Hence my problem.
Just to overcomplicate it and get the concept, how would I go after it using the shell method? It starts off as The
integral from 0 to 2 of 2pi*(x^3+1)...then I lose the shell height. ( P.S., mixing these methods is what stubbed me in the toe. )
On #2...using the definition of the area of a surface, S=Integral of A to B of 2pi*y* sqrt(1+dy/dx)^2), I got
2pi * the integral from 0 to 2 of x^3/9*sqrt(1+(x^2/3)^2) which is equal to 98pi/81.
Thanks guys! Do I need to create a new thread for 3 and 4?