Calculating Volume with Double Integrals: Finding Limits for Enclosed Space

[logicalman]

Hi I am trying to find volume enclosed by following equations:

z = 3x, //Top plane
x^2 + y^2 = 25, // cylinder
x = 4, //line parallel to y axis
x, y=0.


I am trying to figure out what "Limits" should I take on the "Double Integral"
to get the below mentioned Volume ans.

Ans. to this problem is "98 cubic units".

Thanks for any help!

 

[lalbatros]

Hello

I can guess roughly what you are asking.
But your question is not very clear.
I would say that "equations cannot enclose a volume".
I would understand better if you asked the volume of the domain defined from inequalities (I call that constraints).
I guess you ask for the volume of the domain defined by:

$$z < 3x$$​

$$x^2 + y^2 < 25$$​

$$x < 4$$​

$$x>0$$​

$$y>0$$​

However, this volume is infinite. Indeed, negative values for z are allowed.
Maybe you forget the following additional constraint (or something similar):

$$z>0$$​

If this is the case, then I would suggest you to integrate first over the z variable, then on x and y. The integration over the z variable would simply give you the height of the domain for a given (x,y) position in the horizontal plane. This is easy because the boundary of the domain is parallel to the z axis.

Good luck.

 

[logicalman]

"This is the problem for those who can "visualize" particularly a cylinder and plane and also have "sound" knowledge of "double integrals"

Below is the way I am trying to find the volume enclosed by the given equations, but I am not getting the answer!

$$\int (0\ to\ 4)\ dx \int (0\ to \sqrt {x^2 - 25} ) 3x dy$$

(0 to 4) and (o to sqrt x^2 - 25) >definite integral<

nothing is > its "=" to sign everywhere.

I want the Volume formed by the area enclosed by

$$x=4 \ and \ x^2+y^2=25$$

AND the top part
$$z = 3x$$

$$"This is the problem for those who can "visualize" particularly a cylinder and plane and also "sound" knowledge of "double integrals"$$

 

[lalbatros]

Hello

I attached a drawing of the integration domain projected on the x-y plane.
You can see that the easiest way is to:

- integrate along z (since the limits depend only on x on all the domain)
- then along y (since the limits depend only on x on all the domain)
- and terminate by integrating along x.

I write what has to be done:

$$\int_0^4dx \int_0^{\sqrt{5^2-x^2}}dy\int_0^{3x}dz$$​

I think you should obtain the correct result.

Note: your original post did not mention the lower boundary on z (z>0).

 

[logicalman]

$$\int_0^4 3x \sqrt{5^2-x^2}dx$$

I am stuck at this integral. How should I proceed?

if I substitute $$x = 5\sin\theta$$ then I'll have problem with taking the "limits". It will be $$4 = 5\sin\theta$$.

Is there any other way of solving this integral without "substituting"?

 

[logicalman]

By the way that jpg image that you attached is really good!

Which software are you using? :)

 

[lalbatros]

Hello

I would recommend you to persevere with your substitution.
It can be inverted to get the lower and upper limit of the integral, since:

$$\theta = sin^{-1}\left(\frac{x}{5}\right)$$​

By expanding, the square root will become a cosine term.
The differential will produce another cosine term.
Finally, you will have to find the integral of something like:

$$\int \sin\theta \; \cos^2\theta \; d\theta$$​

This integral can be solved again by a subtitution like:

$$new variable = \cos\theta$$​

Take your time to keep track of all the substitutions!

Good luck.

Note1: Another way to solve the integral is to take the argument of the square root as the new variable. You will not need an additional substitution later. Try both ways!

Note2: For the picture, I did it with the MS-Excel drawing tools. I converted it as jpg with the MS-PhotoEditor.

 

[logicalman]

Thanks for the efforts!

here's a simpler way I got from somewhere...

$$u^2 = 5^2 - x^2$$ :tongue:

 

[lalbatros]

Oops

$$\int sin \theta \; cos^2 \theta \; d\theta = -\frac{cos^3\theta}{3}$$

simply from the chain rule for derivatives.

 

[quetzalcoatl9]

lalbatros,

you are correct.

i think that i drank too much coffee

 

[HallsofIvy]

Excuse me, but if the original integral was $\int_0^4 3x \sqrt{5^2-x^2}dx$, wouldn't
"u= 25-x²" be a lot simpler?

 

[logicalman]

Yeah right it will be simpler.