Help with volumes of solids of revolution

[trader]

There are 4 questions I have about calculus. I would greatly appreciate any help on any of them you know.

14. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

y = 2x², y = 0, x = 2

a - the y-axis
b - the x-axis
c - line y = 8
d - line x = 2

18. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 4.

y = x², y = 4

22. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 6.

y = 6 - x, y = 0, y = 4, x = 0

52. Find the volume of the solid whose base is bounded by the graphs of y = x + 1 and y = x² - 1, with the indicated cross sections taken perpendicular to the x-axis.

a. Squares.

b. Equilateral triangles.

I know there are a lot there, but whatever help would be great. I have a test tomorrow so I put 1 problem from each sections--explanation would be awesome!

 

[trader]

Here are the answers I got, but I'm almost sure they are wrong. I tried on my own, but am completely lost.

14
a- 16 pi
b- 25.6 pi
c- 496 pi
d- 17.6 pi

18.
56 pi

22.
36 pi

52.
a- 21.6
b- 5.4

 

[galactus]

Re: Help with volumes Please~!!

52. Find the volume of the solid whose base is bounded by the graphs of y=x+1 and y=x^2-1, with the indicated cross sections taken perpendicular to the x-axis.

a. Squares.

b. Equilateral triangles.

\(\displaystyle \text{I'm sure you know how to find the area of a square. This is like that\\only you're using your functions\\

Find the limits of integration by setting the functions equal\\ to one another and solving for x:}\)

\(\displaystyle x+1=x^{2}-1\)

\(\displaystyle \text{This gives us -1 and 2 as our limits of integration.}\)

So, we have \(\displaystyle \L\\(x+1)-(x^{2}-1)=-x^{2}+x+2\)

\(\displaystyle \text{Now, since the cross-sections are squares, \\we must square our function because it is one side of the square. See?.}\)

\(\displaystyle \L\\\int_{-1}^{2}(x^{2}-x-2)^{2}dx\)


\(\displaystyle \text{Now, for the equilateral triangle, the premise is the same only\\ use the area of an equailateral triangle, instead of a square}\)

\(\displaystyle \text{The area of an equilateral traingle is given by} \frac{\sqrt{3}}{4}s^{2}\)

\(\displaystyle \text{Where s is a side of the triangle, in this case, the area bounded by your functions}\)

\(\displaystyle \L\\\int_{-1}^{2}\frac{\sqrt{3}}{4}(x^{2}-x-2)^{2}dx\)

\(\displaystyle \text{Here's a graph of your region\\Picture that shaded region being a side of your square or triangle and stack the squares or\\
triangles up along the x-axis}\)

 

[trader]

oh wow

thank you---that is extremely helpful!!!


does anyone know about the others ?

 

[soroban]

Hello, trader!

Here's the set-up for #18 . . .

18. Find the volume of the solid generated by revolving the region
bounded by \(\displaystyle y\,=\,x^2,\;y\,=\,4\) about the line \(\displaystyle y\,=\,4\)

                 4|
        - * - - - + - - - * -
           :::::::|:::|::::
           *::::::|:::|r:*:
            *:::::|:::|:* :
              *:::|:::*   :
      -----------***------+-----
                  |       2

We will use "discs": \(\displaystyle \L\,V\;=\;\pi\int^{\;\;\;b}_a(\text{radius})^2\,dx\)

In this problem: \(\displaystyle \,r\:=\:4\,-\,y\:=\:4\,-\,x^2\)

Therefore: \(\displaystyle \L\,V\;=\;\pi\int^{\;\;\;2}_{-2}(4\,-\,x^2)^2\,dx\)

 

[galactus]

18. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 4.

y = x², y = 4

\(\displaystyle \text{I'll help along with one more.\\ See if you can apply the method to the others. \\Give them another try and repost with your attempts. \\We'll see if we can spot the problem}\)

\(\displaystyle \text{Let's do this one with washers and shells \\and see if we get the same answer.}\)

\(\displaystyle \text{When you use shells, the cross-sections are parallel to the line about which you are rotating, \\in this case y=4. If they're parallel to y=4, then they are perpendicular to the y-axis. \\Therefore, we can write our integral in terms of y. See?. }\)

\(\displaystyle \L\\4{\pi}\int_{0}^{4}(y-4)\sqrt{y}dy\)

\(\displaystyle \text{You're probably wondering why the 4pi?.\\ That's because of the square root.}\)

\(\displaystyle \text{Now, washers}\)

\(\displaystyle \L\\{\pi}\int_{-2}^{2}(4-x^{2})^{2}dx\)

\(\displaystyle \text{Perform the integrations on both and see if you get the same thing.\\ It's a good check, if you can do it. Do you see what's going on?.\\
If you are having trouble picturing the situation and setting up your\\integral, I would suggest seeing your teacher and practice, practice.}\)

Edit: uh-oh...Soroban beat me to it. Good to see we jive, so something must be right.

 

[trader]

for 18, my friend said it would be from 0 to 2, not -2 to 2

thats the only conflicting part

 

[galactus]

Your friend forgot about \(\displaystyle (-2)^{2}=4\)