Iterated integral- stuck in the middle

[Sajjad Bin Samad]

Evaluate the following:-

[MATH]\int _0^4\int _{-\sqrt{4x-x^2}}^{\sqrt{4x-x^2}}\:\:\left(x^2+y^2\right)dy\:dx[/MATH]
Since the function is even,

[MATH]2\int _0^4\left[\int _0^{\sqrt{4x-x^2}}\:\:\left(x^2+y^2\right)dy\:\right]dx[/MATH]
After this integrating with respect to y and evaluating at y=0 to y= [MATH]\sqrt{4x-x^2}[/MATH]
we get,

[MATH]\frac{2}{3}\int _0^4\left(\sqrt{4x-x^2}\right)\left(3x^2+4x-x^2\right)\:dx[/MATH]
Beyond this, I can not follow up with my textbook. But what I see in the book is that first, they convert the coordinates to polar ( don't know why or how ) then they use beta and gamma function and got the answer which is 25pi.

Step by step solution is appreciated and also suggest me a book which is very thorough.

 

[MarkFL]

Just to confirm:

[MATH]\int_0^{\sqrt{4x-x^2}} x^2+y^2\,dy=\left[x^2y+\frac{y^3}{3}\right]_0^{\sqrt{4x-x^2}}=\frac{1}{3}\left(3x^2\sqrt{4x-x^2}+(4x-x^2)\sqrt{4x-x^2}\right)=\frac{2x\sqrt{4x-x^2}}{3}(x+2)[/MATH]
And so we now have:

[MATH]I=\frac{2}{3}\int_0^4 2x\sqrt{4x-x^2}(x+2)\,dx[/MATH]
Let's write this as:

[MATH]I=\frac{2}{3}\int_0^4 2x\sqrt{4-(x-2)^2}(x+2)\,dx[/MATH]
Let:

[MATH]w=x-2\implies dw=dx[/MATH]
[MATH]I=\frac{4}{3}\int_{-2}^2 (w+2)\sqrt{4-w^2}(w+4)\,dw[/MATH]
Using the odd-function rule and then the even function rule, this becomes:

[MATH]I=\frac{8}{3}\int_{0}^2 (w^2+8)\sqrt{4-w^2}\,dw[/MATH]
Using an integral table, we may then write:

[MATH]I=\frac{8}{3}\left[\frac{w}{8}(2w^2-4)\sqrt{4-w^2}+2\arcsin\left(\frac{w}{2}\right)+8\left(\frac{w}{2}\sqrt{4-w^2}+2\arcsin\left(\frac{w}{2}\right)\right)\right]_0^2[/MATH]
[MATH]I=\frac{8}{3}\left[\frac{w}{4}(w^2+14)\sqrt{4-w^2}+18\arcsin\left(\frac{w}{2}\right)\right]_0^2=24\pi[/MATH]

 

[Steven G]

I do not think that this needs to be done using tables.

After you brilliantly dealt with the odd/even function I would proceed differently.

[MATH]I=\frac{8}{3}\int(w^2+8)\sqrt{4-w^2}\,dw[/MATH].

I would let \(\displaystyle 2sin(\theta)=\sqrt{4-w^2}.\)

Then \(\displaystyle 2cos(\theta)=w.\)

\(\displaystyle 4 cos^2(\theta) + 8 = w^2 + 8\)

\(\displaystyle -2sin(\theta)d\theta = dw \)

Then [MATH]I=\frac{8}{3}\int(w^2+8)\sqrt{4-w^2}\,dw = \frac{8}{3}\int (4cos^2(\theta)+8)(2sin(\theta))(-2sin(\theta))d\theta[/MATH] ...

 

[Sajjad Bin Samad]

Just to confirm:

[MATH]\int_0^{\sqrt{4x-x^2}} x^2+y^2\,dy=\left[x^2y+\frac{y^3}{3}\right]_0^{\sqrt{4x-x^2}}=\frac{1}{3}\left(3x^2\sqrt{4x-x^2}+(4x-x^2)\sqrt{4x-x^2}\right)=\frac{2x\sqrt{4x-x^2}}{3}(x+2)[/MATH]
And so we now have:

[MATH]I=\frac{2}{3}\int_0^4 2x\sqrt{4x-x^2}(x+2)\,dx[/MATH]
Let's write this as:

[MATH]I=\frac{2}{3}\int_0^4 2x\sqrt{4-(x-2)^2}(x+2)\,dx[/MATH]
Let:

[MATH]w=x-2\implies dw=dx[/MATH]
[MATH]I=\frac{4}{3}\int_{-2}^2 (w+2)\sqrt{4-w^2}(w+4)\,dw[/MATH]
Using the odd-function rule and then the even function rule, this becomes:

[MATH]I=\frac{8}{3}\int_{0}^2 (w^2+8)\sqrt{4-w^2}\,dw[/MATH]
Using an integral table, we may then write:

[MATH]I=\frac{8}{3}\left[\frac{w}{8}(2w^2-4)\sqrt{4-w^2}+2\arcsin\left(\frac{w}{2}\right)+8\left(\frac{w}{2}\sqrt{4-w^2}+2\arcsin\left(\frac{w}{2}\right)\right)\right]_0^2[/MATH]
[MATH]I=\frac{8}{3}\left[\frac{w}{4}(w^2+14)\sqrt{4-w^2}+18\arcsin\left(\frac{w}{2}\right)\right]_0^2=24\pi[/MATH]

Thank you for taking the time to write. I truly appreciate your answer but could you please explain "how and why convert coordinates"?
And please do suggest me a few books as I am doing my Calculus III course (you can even suggest introductory calculus books as I think I need some refreshers). I need thorough explanation of problems so please suggest a book that discusses topics very thoroughly.

 

[Sajjad Bin Samad]

I do not think that this needs to be done using tables.

After you brilliantly dealt with the odd/even function I would proceed differently.

[MATH]I=\frac{8}{3}\int(w^2+8)\sqrt{4-w^2}\,dw[/MATH].

I would let \(\displaystyle 2sin(\theta)=\sqrt{4-w^2}.\)

Then \(\displaystyle 2cos(\theta)=w.\)

\(\displaystyle 4 cos^2(\theta) + 8 = w^2 + 8\)

\(\displaystyle -2sin(\theta)d\theta = dw \)

Then [MATH]I=\frac{8}{3}\int(w^2+8)\sqrt{4-w^2}\,dw = \frac{8}{3}\int (4cos^2(\theta)+8)(2sin(\theta))(-2sin(\theta))d\theta[/MATH] ...

Thank you for your assistance with this problem and I really appreciate your idea of not using an integral table. But I am still confused about converting coordinates. Please suggest me some books that are very through starting from Calculus I to III.

 

[MarkFL]

Thank you for taking the time to write. I truly appreciate your answer but could you please explain "how and why convert coordinates"?
And please do suggest me a few books as I am doing my Calculus III course (you can even suggest introductory calculus books as I think I need some refreshers). I need thorough explanation of problems so please suggest a book that discusses topics very thoroughly.

I was simply trying to show you how I would do it without using polar coordinates, and certainly could be done without a table (if you find that troublesome) at the end, likely using IBP. I'm not a reviewer of textbooks so I can't help you there.

 

[Li Batton]

Converting to polar coordinates makes this "MUCH" easier to evaluate.

First lets write down how to convert Cartesian coordinates to polar

[MATH]x={\it rcos} \left( \theta \right) [/MATH]
[MATH]y={\it rsin} \left( \theta \right) [/MATH]
[MATH]{r}^{2}={x}^{2}+{y}^{2}[/MATH]
Now let us look at the bounds of the inner integral
To make things a little easier let us focus on the positive
square root.

[MATH]y=\sqrt {4\,x-{x}^{2}}[/MATH]
If we square both sides and complete the square this becomes we
a circle centered at (2,0)

[MATH] \left( x-2 \right) ^{2}+{y}^{2}=4[/MATH]
Now we need to find how how r in polar coordinates varies with an
angle theta. In other words we need to find the equation of the circle in polar coordinates.

[MATH] \left( {\it rcos} \left( \theta \right) -2 \right) ^{2}+ \left( {\it rsin} \left( \theta \right) \right) ^{2}=4 [/MATH]
If we expand and simplify we get

[MATH]r=4\,\cos \left( \theta \right) [/MATH]
This is the equation of the same circle in polar coordinates
So r will vary between zero and [MATH]4\,\cos \left( \theta \right) [/MATH]
Now the angle theta will trace out the whole circle counter clockwise as
theta goes from o to pi.

The double integral in polar coordinates is
[MATH]\int_{0}^{\pi }\int_{0}^{4cos(\theta )}(r^2)rdrd\theta[/MATH]
The extra r has to be there because our differential area is

[MATH]dA=rdrd\theta [/MATH]
So we have

[MATH]\int_{0}^{\pi}\int_{0}^{4cos(\theta )}r^3drd\theta [/MATH]
Which will reduce to

[MATH]64\int_{0}^{\pi}cos^4\theta d\theta =24\pi [/MATH]
I will leave the evaluation of the last integral to you.
Just use power reduction trig identity

I hope this helps you

 

[Steven G]

Thank you for your assistance with this problem and I really appreciate your idea of not using an integral table. But I am still confused about converting coordinates. Please suggest me some books that are very through starting from Calculus I to III.

The converting of integrals from rectangular to polar coordinates is a standard procedure shown in every calculus book which I have seen.
That is any standard calculus book will have this.
\(\displaystyle x=rcos(\theta)\)
\(\displaystyle y=rsin(\theta)\)
\(\displaystyle dxdy=rdrd\theta\)

 

[Sajjad Bin Samad]

Converting to polar coordinates makes this "MUCH" easier to evaluate.

First let us write down how to convert Cartesian coordinates to polar

[MATH]x={\it rcos} \left( \theta \right) [/MATH]
[MATH]y={\it rsin} \left( \theta \right) [/MATH]
[MATH]{r}^{2}={x}^{2}+{y}^{2}[/MATH]
Now let us look at the bounds of the inner integral
To make things a little easier let us focus on the positive
square root.

[MATH]y=\sqrt {4\,x-{x}^{2}}[/MATH]
If we square both sides and complete the square this becomes we
a circle centered at (2,0)

[MATH] \left( x-2 \right) ^{2}+{y}^{2}=4[/MATH]
Now we need to find how r in polar coordinates varies with an
angle theta. In other words, we need to find the equation of the circle in polar coordinates.

[MATH] \left( {\it rcos} \left( \theta \right) -2 \right) ^{2}+ \left( {\it rsin} \left( \theta \right) \right) ^{2}=4 [/MATH]
If we expand and simplify we get

[MATH]r=4\,\cos \left( \theta \right) [/MATH]
This is the equation of the same circle in polar coordinates
So r will vary between zero and [MATH]4\,\cos \left( \theta \right) [/MATH]
Now the angle theta will trace out the whole circle counter-clockwise as
theta goes from o to pi.

The double integral in polar coordinates is
[MATH]\int_{0}^{\pi }\int_{0}^{4cos(\theta )}(r^2)rdrd\theta[/MATH]
The extra r has to be there because our differential area is

[MATH]dA=rdrd\theta [/MATH]
So we have

[MATH]\int_{0}^{\pi}\int_{0}^{4cos(\theta )}r^3drd\theta [/MATH]
Which will reduce to

[MATH]64\int_{0}^{\pi}cos^4\theta d\theta =24\pi [/MATH]
I will leave the evaluation of the last integral to you.
Just use power reduction trig identity

I hope this helps you

Thank you very much and I really appreciate your clear and thorough explanation.

 

[Steven G]

Thank you for your assistance with this problem and I really appreciate your idea of not using an integral table. But I am still confused about converting coordinates. Please suggest me some books that are very through starting from Calculus I to III.

I thought initially that you wanted a book that covers converting coordinates, but I was wrong.
A thorough book for the calculus sequence is one by Thomas and Finney (not Finney and Thomas !)
Here is a link to the edition I have.

 

[Sajjad Bin Samad]

I thought initially that you wanted a book that covers converting coordinates, but I was wrong.
A thorough book for the calculus sequence is one by Thomas and Finney (not Finney and Thomas !)
Here is a link to the edition I have.

Thanks a lot and ah one more thing- what about this one or this one.

 

[Steven G]

Thanks a lot and ah one more thing- what about this one or this one.

Anton is a popular author. I am sure that he does a good job. If you really want a rigorous book then the one I recommended is the best. Next to that book, Anton will be just fine.