Maths question [MTH1020]

[alondouek]

So I have a question;

Find   using trigonometric substitution.

I got , which I think is right, but I'm not too sure if it is or how I got there lol

Can someone please help?

[b^3]

We want to turn the into something we can work with, so we will make use of the trigonometric identity , which becomes . To make the inside of the square root look like this we will try a substitution. Lets try .

Spoiler

Which gives the same answer you had.

[alondouek]

Thanks b^3, really appreciate it!

Why exactly are we substituting ? Would it be equally valid to use , as ?

(I'm assuming the 3 is there because of the 9 in the square-root expression? I don't really have a great handle on this theory)

[b^3]

should work as well (just be careful of the negative from the derivative). We have the there because we want to form "", if we just substituted in , then we'd end up with , which is not what we want as it doesn't satisfy our trigonometric identity. But what do we have to do to satisfy it? We need to get a factor of out the front of the , so that we can take a factor of out of both of the terms, leaving , which is exactly what we want. We can now turn that into our .

EDIT: Had sines and cosines flipped around, fixed now.

[alondouek]

Thanks again, great explanation

[alondouek]

So I was thinking about this over the weekend, and I'm still kind of stumped by it.

How do we know which trigonometric expression to substitute? Is there a defined method of selecting one, or is it just something you ascertain from what you're substituting?

tbh I'm finding trig substitution as a whole quite puzzling

[brightsky]

just think logically, and try stuff out. say you got an integrand sqrt(a^2-x^2). now won't it be nice if you could get rid of that square root? to do that, we need some square number under the square root. how do we obtain that? use a trig identity! which one? well the expression looks roughly like 1-something^2. ring any bells? use identity cos^2 (x) = 1-sin^2(x) or the like.

trig substitution doesn't always work though. when you get more complicated integrands, you'll just have to think about all the integration techniques in your toolbox and try stuff out.

[b^3]

You want to pick a substitution that changes the square root into something that you can work with. We make use of the relevant trigonometric identity to get the square root of a trig squared, which gives us a mod. Then normally we'd be integrating over a domain where the sign of this mod doesn't change, allowing us to remove the mod and pick the appropriate sign.
It pretty much boils down to this in the end.

EDIT: Beaten, why'd I decided to spend time centering the image

[alondouek]

Amazing! Huge thanks to both of you

[lzxnl]

So I have a question;

Find   using trigonometric substitution.

I got , which I think is right, but I'm not too sure if it is or how I got there lol

Can someone please help?

I have an easier way of doing that.
We want the area of half a semicircle as your function is that of a semicircle, while the integration bounds span half the domain. So you just have pi*r^2/4 = 9pi/4
Is this considered cheating?

Alternatively you CAN do this integral by parts...although that might be overkill for this question.

[alondouek]

Don't know if anyone is still awake, but how can I tell when to use either the cylindrical-shell method or the disk method to find a volume by integration?

[BigAl]

You can use a right angled triangle to figure out the identities. I didn't do mth1020 but from what I've seen on YouTube, you often see fractions in this kind of problems...check prof burgers videos on YouTube. They are the best for introductory calculus

[brightsky]

Don't know if anyone is still awake, but how can I tell when to use either the cylindrical-shell method or the disk method to find a volume by integration?

shell method is used when the terminals are parallel to the axis about which you are rotating. disk method is used when the terminals are perpendicular to the axis about which you are rotating.

for example, say you want to calculate the volume of the solid produced when you rotate the area bounded by lnx, x = 2, x=4 and the x-axis about the y-axis. do you see how the terminals x=2 and x=4 are parallel to the y-axis (the axis about which you are rotating)? in cases like this, use shell integration. however, if you wanted to calculate the volume of the solid produced when you rotate this area about the x-axis, you would use disk method, as per usual. do you see how the terminals x= 2 and x=4 are perpendicular to the x-axis?

[alondouek]

Thanks brightsky!

So if I'm rotating around the x-axis the function y=x^3, enclosed by y=1, then I should use the shell method as y=1 is parallel to the axis of rotation?

[lzxnl]

Thanks brisghtsky!

So if I'm rotating around the x-axis the function y=x^3, enclosed by y=1, then I should use the shell method as y=1 is parallel to the axis of rotation?

That's not quite enough info...what else are you given? You can't just have the one bound.