1. Apr 8, 2005

### 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! :tongue2:

2. Apr 8, 2005

### 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?

3. Apr 8, 2005

### ehild

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

$$dV= r d \phi dr dz$$

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

$$0 \le \phi \le 2\pi$$

$$0 \le r \le 2$$

So you have to calculate the integral

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

4. Apr 8, 2005

### 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?

Last edited: Apr 8, 2005
5. Apr 8, 2005

### 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$$}\;dz\;dy\;dx = \frac{4}{\pi^2}$$

6. Apr 8, 2005

### VinnyCee

Anyone double check the last calculation yet?

7. Apr 8, 2005

### ehild

It looks all right.

ehild

8. Apr 8, 2005

### marlon

That is indeed correct

marlon

9. Apr 10, 2005

### VinnyCee

Many thanks

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