MHB Finding volumes by multiple integrals

[aruwin]

How do I solve this? How do I determine the range? Ill they be triple integrals?Please explain to me.

Find the volumes in R³.

1. Find the volume U that is bounded by the cylinder surface x^2+y^2=1 and the plane
surfaces z=2, x+z=1.

2. Find the volume W that is bounded by the cylindrical surfaces
x^2 + y^2 = 1 and x^2 + z^2 = 1.

I know these are 2 questions and they're both about finding volumes but the second one seems to have 2 cylinders http://www.mathhelpboards.com/images/smilies/confused.gif ??I don't get it.

 

[BAdhi]

you can either use triple integrals or double integrals for the first part, but for this question triple integrals is much easier to me.

by definition,

[math]V= \int\int\int_U dV[/math]

[math]dv=dxdydz[/math]

first let's take x as the independent variable, thus limits of z can be decided by x. So the z changes between z=2 plane and z=1-x plane which makes the boundary of the z axis.

Lets take the boundary of the xy plane is S

[math]V=\int\int_S \int_{1-x}^2 dxdydz[/math]

[math]V=\int\int_S \left[z\right]_{1-x}^2 dxdy[/math]

[math]V=\int\int_S 2-1+x dxdy[/math]

[math]V=\int\int_S 1+x dxdy[/math]

now it is double integrals so, you can use polar coordinate system to calculate the integral (since the boundary S is a circle with the equation [math]x^2+y^2=1[/math]).

If you use double integrals, it will also come to the above equation.

For that you have to find the double integrals of two equations (since double integral finds the volume between a surface and z=0 plane), i.e.

[math]V_1= \int\int_S 2 dxdy[/math]
[math]V_2=\int\int_S 1-x dxdy[/math]

since in the S region (bounded by [math] x^2 +y^2=1[/math]) z=2 is above respect to z=1-x to find the volume [math]V[/math]

[math]V=V_1-V_2[/math]

[math]V=\int\int_S 2 dxdy - \int\int_S 1-xdxdy[/math]

[math]V=\int\int_S 2-(1-x) dxdy[/math]

[math]V=\int\int_S 1+xdxdy[/math]

I guess you can continue from here :)

 

[Sudharaka]

Hi BAdhi, :)

Just want to point out some small mistakes/suggestions you may have overlooked. :)

... Let's take the boundary of the xy plane is S ...

In fact you have taken \(S\) to be the projection of the volume \(U\), on the xy-plane, which is a disk of radius 1.

... \(\displaystyle V=\int\int_S \int_{1-x}^2 dxdydz\) ...

I think this should be, \(\displaystyle V=\int\int_S \int_{1-x}^2 dz\,dx\,dy\).

...

If you use double integrals, it will also come to the above equation.

For that you have to find the double integrals of two equations (since double integral finds the volume between a surface and z=0 plane), i.e.

[math]V_1= \int\int_S 2 dxdy[/math]
[math]V_2=\int\int_S 1-x dxdy[/math]

since in the S region (bounded by [math] x^2 +y^2=1[/math]) z=2 is above respect to z=1-x to find the volume [math]V[/math]

[math]V=V_1-V_2[/math]

[math]V=\int\int_S 2 dxdy - \int\int_S 1-xdxdy[/math]

[math]V=\int\int_S 2-(1-x) dxdy[/math]

[math]V=\int\int_S 1+xdxdy[/math]

.....

This is correct, however by taking a vertical strip of the volume \(U\) we can get the double integral directly. The length of such a strip would be, \(2-(1-x)=1+x\). Therefore the volume of the region could be written as,

\[\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}(1+x)dx\,dy\]

Kind Regards,
Sudharaka.-------------------------------------------------------------------------------------------------------------------------------------------------------

How do I solve this? How do I determine the range? Ill they be triple integrals?Please explain to me.

Find the volumes in R3.

1. Find the volume U that is bounded by the cylinder surface x^2+y^2=1 and the plane
surfaces z=2, x+z=1.

2. Find the volume W that is bounded by the cylindrical surfaces
x^2 + y^2 = 1 and x^2 + z^2 = 1.

I know these are 2 questions and they're both about finding volumes but the second one seems to have 2 cylinders ??I don't get it.

Hi aruwin, :)

You will have to set up triple integrals to calculate the volumes of these regions. As you may know, for the calculation of volumes, the function of integration is the constant function 1 (See this). The method of solving triple integrals is informally known as the "shadow method", and I suggest you to watch this to get an idea about this method. Also this would be helpful.

As an example, I shall explain how to calculate the volume of the second region, as you have said it is the part that you don't understand amongst the two questions.

Our region is bounded by the curves,

\[x^2+y^2=1\mbox{ and }x^2+z^2=1\]

Always draw figures of the given surfaces, and visualize the region that you have to find the volume. Then you will be able to find the boundaries of integration easily.

It could be seen that, \(z\) varies between \(-1\mbox{ and }1\). That is,

\[-1\leq z\leq 1\]

The projection(a shadow casted by a light source coming form the positive z direction) of the region on the xy-plane would then be a circle with radius 1. Therefore the triple integral that gives the volume of the region could be written as,

\[\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{-1}^{1}dz\,dx\,dy=4\int_{-1}^{1}\sqrt{1-y^2}dy\]

Evaluation of the integral will yield,

\[\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{-1}^{1}dz\,dx\,dy=2\pi\]

Kind Regards,
Sudharaka.