# root expression

#### DBA

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.

#### Anonymous1

I am having trouble to rewrite this expression:
Where exactly do you want to get?

Some stuff:

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

$$\displaystyle 5\sqrt{4-y^2}-y\sqrt{4-y^2}$$

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

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

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

• DBA

#### DBA

Thanks. Actually I need to get some expression I can integrate but I forgot the algebra part of it...

#### wonderboy1953

Suggestion

Multiply out what you have in the parentheses would be the first step.

• DBA

#### Anonymous1

Multiply out what you have in the parentheses would be the first step.
Check out the latex. The square root is only over the first parenthesis.

So...

#### AllanCuz

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,

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