Easy calculus questions

[Schadenfroh]

I have let my math skills slack while focusing on my chemistry (organic chemistry is a BIG time sink, but it makes a hell of a lot more sense than Calculus), I am stuck on these two problems (studying for an exam, finding questions in the book that I cannot do just in case he asks something similar), these two are driving me mad and I have worked on them for the past hour with little progress.

S = anti derivative symbol
S Pi * r^2 dx = formula for finding the volume of these things flipped across the axis.

1st off:

Find the volume of 3D objects below:

The region in the first quadrant bounded above by the line y = 2, below by the curve y = 2Sinx, 0 <= X <= Pi / 2, and on the left by the y-axis, about the line y = 2.

What I have done thus far (using S PiR^2:


Pi * S (2 - 2sinx)^2 dx = 2*Pi S (1-sinX)^2 dx = 2PI S (1 - 2SinX - (SinX)^2)

This is all fine and dandy, but I need to get the anti derivative of (sinX)^2. I cant use substitution here (it is not multiplied by Cosx) and I that the antiderivative of -(SinX)^2 is not (CosX)^2. Trig identities are useless, for all I could convert SinX into using (using the ident of (SinX)^2 + (CosX)^2 = 1) would just create another trig function that I cannot take the anti derivative of (CosX)^2. How in the world do I find the anti derivative of Sin^2? I need that integral so I can integrate it into 0 to Pi / 2 and find the volume.

Also, this one has been giving me trouble as well, as I cannot recall the specific rules:

Y = sinx*cosx between the x axis

I have to use the formula for rotating it across the axis, so it comes to this:

(here is what I have so far)
Pi S (sinx*cosx)^2 = Pi S (Sin^2)(Cos^2).

I tried substitution, u = sin, du = cos, but du would have to be squared if that were true, therefore I cannot substitute. Again I am struck with waht to do with Sin^2 when I want to get the anti derivative of it. I know I am making a simple mistake somewhere in there and once I get the anti derivative, it will be clear sailing to integration.

help......

thanks

 

[Eeezee]

You're taking the integral of (sinx)^2? Something tells me that you're not doing this problem right, because integrating (sinx)^2 is 1000x more complicated than what you're learning.

Also, you pulled out that 2 incorrectly. 2^2 = 4, not 2

Specifically, what you need to know is that integrating an equation gives the area under its curve. (I went ahead and read what you were supposed to do after I wrote the above 3 sentences). Your integral should be
S (2-2sinx)dx

That gives you the area enclosed by those 2 lines, as you already realize. You're looking for the volume under a rotation about the Z-axis. Are you certain that's the right formula for finding this sort of answer?

 

[TuxDave]

Edit: Well... actually after giving it a try I got stuck in a stupid loop..

 

[Schadenfroh]

Formula I am using (straight from the prof.): V = S Pi[R(x)]^2 dx

I moved the two outside the problem to simplify it, if I did not have to square the dang thing it would be a hell of a lot easier. My prof will not be there tomorrow and this is going to drive me mad over the weekend.

 

[jersiq]

Originally posted by: Schadenfroh
Formula I am using (straight from the prof.): V = S Pi[R(x)]^2 dx

I moved the two outside the problem to simplify it, if I did not have to square the dang thing it would be a hell of a lot easier. My prof will not be there tomorrow and this is going to drive me mad over the weekend.

Doesn't seem like the correct forms for evaluating a solid of revolution.
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