root expression

DBA

May 2009
129
7
I am having trouble to rewrite this expression:

\(\displaystyle sqrt(4-y^2)(5-y)\)

I would start with \(\displaystyle 5*sqrt(4-y^2) (-y)*sqrt(4-y^2)\)
But then I am stuck.
Thank.
 
Nov 2009
517
130
Big Red, NY
I am having trouble to rewrite this expression:
Where exactly do you want to get?

Some stuff:

\(\displaystyle \sqrt{4-y^2}\cdot(5-y)\)

\(\displaystyle 5\sqrt{4-y^2}-y\sqrt{4-y^2}\)

\(\displaystyle \sqrt{25(4-y^2)}-\sqrt{y^2(4-y^2)}\)

\(\displaystyle \sqrt{100-(5y)^2}-\sqrt{4y^2-(y^2)^2}\)

\(\displaystyle \sqrt{(10-5y)(10+5y)}-\sqrt{(2y+y^2)(2y-y^2)}\)
 
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DBA

May 2009
129
7
Thanks. Actually I need to get some expression I can integrate but I forgot the algebra part of it...
 
Oct 2009
769
87
Suggestion

Multiply out what you have in the parentheses would be the first step.
 
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Apr 2010
384
153
Canada
Thanks. Actually I need to get some expression I can integrate but I forgot the algebra part of it...
These are very different questions. Anon gave a great way to do a whole bunch of things to that equation...but do we want to integrate that? Yikes! I don't want to.

\(\displaystyle \sqrt{4-y^2}(5-y) = 5 \sqrt{4-y^2} - y \sqrt{4-y^2} \)

If we have an integral it would look like

\(\displaystyle \int \sqrt{4-y^2}(5-y) dy = 5 \int \sqrt{4-y^2}dy - \int y \sqrt{4-y^2} dy \)

Let us go through both parts here.

Part 1

\(\displaystyle 5 \int \sqrt{4-y^2}dy\)

This is of the form \(\displaystyle \sqrt{a^2 - y^2} \) which is a prime candidate for sin substitution (I will discuss a trick and why we dont need to do this in a moment!)

Let \(\displaystyle y=2sin \theta \) and \(\displaystyle dy= 2cos \theta d \theta \)

\(\displaystyle 5 \int \sqrt{4-4sin^2 \theta }dy = 5 \int 2\sqrt{1 - sin^2 \theta } 2 cos \theta d \theta = 20 \int cos^2 d \theta \)

Remember the double angle formula

\(\displaystyle cos^2 \theta = \frac{ 1+cos2 \theta }{2} \)

\(\displaystyle 20 \int cos^2 d \theta = 20 \int \frac{1}{2} d \theta + 10 \int cos2 \theta d \theta = 10 \theta + 5 sin2 \theta \)

Of course you then switch your limits and evaluate for theta. However, if we don't want to do all of this we can note that

\(\displaystyle 5 \int \sqrt{4-y^2}dy\) is 5 times the area of a semi-circle of radius 2. Therefore,


\(\displaystyle 5 \int \sqrt{4-y^2}dy = 5 \frac{ \pi (2)^2 }{2} = 10 \pi\)

Part 2

\(\displaystyle \int y \sqrt{4-y^2} dy \)

In this case let \(\displaystyle p=y^2 \) and \(\displaystyle dp = 2ydy \)

\(\displaystyle \frac{1}{2} \int \sqrt{4-p} dp = - \frac{1}{3} (4-p)^{ \frac{3}{2} } \)

Subbing back in for y,

\(\displaystyle - \frac{1}{3} (4-y^2)^{ \frac{3}{2} } \)

Thus,

\(\displaystyle 5 \int \sqrt{4-y^2}dy - \int y \sqrt{4-y^2} dy = 10 \pi - \frac{1}{3} (4-y^2)^{ \frac{3}{2} } \)