You're right about the integral using dy instead of dx, but you forgot about the squared in the solid of revolution.
Remember that the volume is given by:
Therefore in your case it would be:
so embarrassed that this is my second thread and in both i'm asking for basic help.
i have taken calc 1 but i have not retained the material very well. i'm now in a class that asks us to solve problems using software and i'm a mess.
i wish i could contribute by helping people here, but i am not very advanced at all [EDIT TO ADD: I did finally find some posts to answer, so I feel better ! ]
the problem is as follows:
Find the volume of the solid that is created by evaluating the function cos(x) from –π/2 to π/2 and then rotating the figure about the y axis.
now to show effort towards a solution:
i believe that this would be a cylinder, and possibly dy instead of dx as it rotates around the y axis. the way i'd imagine it is by
π *(integral of x= cos y) from –π/2 to π/2 which comes out to 2π
am i on the right track?
thanks in advance for any hints!
This is a trickier on than it looks
First, rearrange for x=arcos(y), and find the volume generated by rotating the area from terminals x=0 to x=pi/2 i.e y=0 to y=1 around the axis using Educated's integral with f(y)=arcos(y). The area you are rotating from -pi/2 to pi/2 will give twice volume required (try picturing this in your head)
Wow, thanks for this response. I did not translate the function for y.
Do you mean to say that the volume will be twice the result of integrating from y= 0 to y= 1?The area you are rotating from -pi/2 to pi/2 will give twice volume required (try picturing this in your head)
Thats precisely correct
Think about it this way:
You are rotating two of the same area 360 degrees about the axis, this will give you twice the volume
You need only half the total area to be rotated around for the solid required
Sketch the graph, labelling co-ordinates. It helps to visualise and to switch between the axis with dy or dx more easily!!
The integration also looks quite tricky, you probably need extra help on this. I'm not exaclty sure how to integrate arcos^2(y)???
I wondered about that, as I often see integrals from -x to 0 or 0 to x but never -x to x. So your logic makes sense to me.
As for integrating arccos--neither do I , but luckily this is a computing course. I input this into Maple and get:
I tried Maple's Volume of Revolution wizard but it gave me a different answer....However, it wouldnt let me take the integral with respect to y so that may account for the difference.