Correcting Limits in Double Integral for Intersection Volume of Three Cylinders

[Scigatt]

Homework Statement

I was trying to find the volume of the intersection between 3 cylinders x^2 + y^2 = 1, y^2 + z^2 =1, and z^2 + x^2 =1. I set up the double integral in two different ways:
$$8\int_{\theta = \frac{-\pi}{4}}^{\frac{\pi}{4}}\int_{r = 0}^{1} \sqrt{1 - r^{2}\: cos^{2}\, \theta}\; \; r\: dr\: d\theta$$
$$16\int_{\theta = 0}^{\frac{\pi}{4}}\int_{r = 0}^{1} \sqrt{1 - r^{2}\: cos^{2}\, \theta}\; \; r\: dr\: d\theta$$

They should give the same answer, but they don't. Apparently the second one is supposed to be right.

Homework Equations

see above

The Attempt at a Solution

After doing all the integrating, I get
$$\frac{8}{3} \left [ tan\, \theta \; - \; sec\, \theta \;- \; cos\, \theta \right ]_{\frac{-\pi}{4}}^{\frac{\pi}{4}} = \frac{16}{3}$$
$$\frac{16}{3} \left [ tan\, \theta \; - \; sec\, \theta \;- \; cos\, \theta \right ]_{0}^{\frac{\pi}{4}} = 8(2 - \sqrt{2})$$

what's going on here?

 

[Hootenanny]

Homework Statement

I was trying to find the volume of the intersection between 3 cylinders x^2 + y^2 = 1, y^2 + z^2 =1, and z^2 + x^2 =1. I set up the double integral in two different ways:
$$8\int_{\theta = \frac{-\pi}{4}}^{\frac{\pi}{4}}\int_{r = 0}^{1} \sqrt{1 - r^{2}\: cos^{2}\, \theta}\; \; r\: dr\: d\theta$$
$$16\int_{\theta = 0}^{\frac{\pi}{4}}\int_{r = 0}^{1} \sqrt{1 - r^{2}\: cos^{2}\, \theta}\; \; r\: dr\: d\theta$$

They should give the same answer, but they don't. Apparently the second one is supposed to be right.

Homework Equations

see above

The Attempt at a Solution

After doing all the integrating, I get
$$\frac{8}{3} \left [ tan\, \theta \; - \; sec\, \theta \;- \; cos\, \theta \right ]_{\frac{-\pi}{4}}^{\frac{\pi}{4}} = \frac{16}{3}$$
$$\frac{16}{3} \left [ tan\, \theta \; - \; sec\, \theta \;- \; cos\, \theta \right ]_{0}^{\frac{\pi}{4}} = 8(2 - \sqrt{2})$$

what's going on here?

Your second result is correct.

The error with your first result occurs with your integration with respect to z. In the first case, you have integrated r*dz between z = 0 and z= sqrt[1-r^2*cos(theta)]. That is, the region in the first Octant. However, when it comes to your integration of theta, you have integrated over two octants.

 

[Scigatt]

Your second result is correct.

The error with your first result occurs with your integration with respect to z. In the first case, you have integrated r*dz between z = 0 and z= sqrt[1-r^2*cos(theta)]. That is, the region in the first Octant. However, when it comes to your integration of theta, you have integrated over two octants.

Are you talking about octant in the plane or octants in space? I'm not sure how the 'mismatched' octants would lead to such an error. Here is the preliminary triple integral of the first integral:

$$8\int_{\theta = \frac{-\pi}{4}}^{\frac{\pi}{4}}\int_{r = 0}^{1} \int_{z = 0}^{\sqrt{1 - x^{2}}}\: dz\: r\: dr\: d\theta$$

You say that there is a flaw in the in the first integral because of a conflict in the z and theta limits. How would you change the integral without changing the theta-limits so that it gives the right answer?