- #1
Lancelot59
- 646
- 1
I just have a few small questions.
For this first question I need to setup the integral to find the volume defined by.
[tex]y=tan^{3}(x)[/tex], [tex]y=1[/tex], and [tex]x=0[/tex] by rotating it about y=1.
My result was this:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(tan^{3}(x) - 1)^{2}dx}[/tex]
From my understanding all you need to do is offset the function by the axis in order to make it work. But the actual answer is:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(1 - tan^{3}(x))^{2}dx}[/tex]
Which looks like it would do something else entirely. I had a similar issue when doing the same thing for sin(x) about y=1. What's going on here?
My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
For this first question I need to setup the integral to find the volume defined by.
[tex]y=tan^{3}(x)[/tex], [tex]y=1[/tex], and [tex]x=0[/tex] by rotating it about y=1.
My result was this:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(tan^{3}(x) - 1)^{2}dx}[/tex]
From my understanding all you need to do is offset the function by the axis in order to make it work. But the actual answer is:
[tex]\pi\int_{0}^{\frac{\pi}{4}} {(1 - tan^{3}(x))^{2}dx}[/tex]
Which looks like it would do something else entirely. I had a similar issue when doing the same thing for sin(x) about y=1. What's going on here?
My second issue is a conceptual question involving work. Just to make sure I have the order of everything correct, how does say, the work done to move a spring relate to it's potential energy? Is it the integral of work = potential energy?
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