How Do You Set Up Limits for a Triple Integral Involving a Cylinder and Planes?

[VinnyCee]

Ok I hve a triple integral problem for find the area of the following:

cylinder: x^2 + y^2 = 4

plane: z = 0

plane: x + z = 3

It is a cylinder cut at the xy-plane and by the last plane. It looks like a circular wedge standing straight up from the xy-plane.

I just can't figure out the limits of integration, I know the last variable integrated must have limits that are constants. I just don't know what to do! Help me please!

 

[VinnyCee]

Here is what I have so far

By solving the equation x + z = 3 for Z, z = 3 - x I can get the z-limits to be from 0 to 3 - x, right?

I then solve x^2 + y^2 = 4 out to be y = \pm \sqrt{4 - x^2}, and those are the y-limits?

After that, the x-limits are from -2 to 2, right?

Here is that in integral form: \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz dy dx = 12\pi

Am I on the right track here?

 

[ehild]

By solving the equation x + z = 3 for Z, z = 3 - x I can get the z-limits to be from 0 to 3 - x, right?

I then solve x^2 + y^2 = 4 out to be y = \pm \sqrt{4 - x^2}, and those are the y-limits?

After that, the x-limits are from -2 to 2, right?

Here is that in integral form: \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz dy dx = 12\pi

Am I on the right track here?

It looks perfect.

If you use cylindrical coordinates: r, phi, z, the integration becomes easier. In these coordinates

x=\cos\phi
y= \sin \phi
z=z,

the volume element is

dV= r d \phi dr dz

and your boundaries are

0 \le z \le 3-x = 3 - r \ cos\phi

0 \le \phi \le 2\pi

0 \le r \le 2

So you have to calculate the integral

\int_{0}^{2} \int_{0}^{2\pi}} \int_{0}^{3 - r \cos(\phi)}r dz d\phi dr

 

[VinnyCee]

Here is the triple integral solved step-by-step

V = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz\; dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[ z \right]_{0}^{3 - x} dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[(3 - x) - (0)\right] dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} (3 - x) dy\; dx

V = \int_{-2}^{2} \left[3y - xy\right]_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} dx = \int_{-2}^{2} \biggl\{\left[3\left(\sqrt{4 - x^2}\right) - x\left(\sqrt{4 - x^2}\right)\right] - \left[3\left(-\sqrt{4 - x^2}\right) - x\left(-\sqrt{4 - x^2}\right)\right]\biggl\} dx

V = \int_{-2}^{2} \left(6 \sqrt{4 - x^2} - 2x \sqrt{4 - x^2}\right) dx = 6 \int_{-2}^{2} \left(4 - x^2\right)^{\frac{1}{2}}\; dx + \int_{-2}^{2} -2x \sqrt{4 - x^2}\; dx

u = 4 - x^2 and du = -2x dx for that second integral.

V = 6\left[\frac{2}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right)\right]_{-2}^{2} +\; 2\left[\frac{2}{3}\left(4 - x^2\right)^{\frac{3}{2}}\right]_{-2}^{2}

V = 6\biggl\{\left{\sqrt{4 - 4} + 2 \sin^{-1}\left(-1\right)\right] - \left[\sqrt{4 - 4} + 2\sin^{-1}\left(1\right)\right]\biggl\} + 2\biggl\{\left[\frac{2}{3} \left(4 - 2^2\right)\right] - \left[\frac{2}{3} \left(4-(-2)^2\right)\right]\biggl\}

V = 6\left(2\pi\right) + 2\left(0\right) = 12\pi

Does this check out? Should I post a graph?

 

[VinnyCee]

Ok, a new triple integral

Did I do this new Problem correctly? :

Find volume of the region in the first octant bounded by the plane y = 1 - x and the surface z = \cos\left(\frac{\pi\;x}{2}\right). And 0\;\le x \;\le1.

Integral:

\int_{0}^{1} \int_{0}}^{1 - x} \int_{0}^{\cos\bigl\(\frac{\pi\;x}{2}}\bigl\)}\;dz\;dy\;dx = \frac{4}{\pi^2}

 

[VinnyCee]

Anyone double check the last calculation yet?

 

[ehild]

Anyone double check the last calculation yet?

It looks all right.

ehild

 

[marlon]

That is indeed correct

marlon

 

[VinnyCee]

Many thanks

Thank you ehild and marlon. I am never sure of some of these calc problems!