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I was asked to find the volume of the shape that is the intersection of the two cylinders $x^2+y^2 \leq 1$ and $x^2+z^2 \leq 1$

My solution is false, I would like to know why it is not good.

by reducing 1 equations from the other, you can get $y^2-z^2 \leq 1$, and so $y^2 \leq 1+z^2$

so the bounds for $y$ are: $-\sqrt{1+z^2} \leq y \leq \sqrt{1+z^2}$

since $x^2+z^2 \leq 1$ we can get that $z^2 \leq 1-x^2$ and as such $-\sqrt{1-x^2} \leq z \leq \sqrt{1-x^2}$

and $-1 \leq x \leq 1$ is our "free" variable that has no restrictions.

using wolfram alpha, i calculate this tripe integral which should theoretically give me the volume of the shape requested:

$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{-\sqrt{1+z^2}}^{\sqrt{1+z^2}}1dydzdx$$ and the result wolfram alpha gave me is $6.99215$.

After googling a bit, I stumble upon this page http://www.math.tamu.edu/~Tom.Kiffe/calc3/newcylinder/2cylinder.html that says the answer is $\frac{16}{3}$.

1) I do not understand how they calculated the volume on that page.

2) I don't understand where I went wrong.

Would appreciate clarification.

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The first two $\iint$ represent your region of integration here you are using $\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} $ this is a very well known region, it's circular region.

In method $(2)$ When you intersect two cylinders, you get a square region as in the link.

enter image description here

The also note that the height function of this intersection of cylinders are different on different regions, as one cylinder's height is taller than other on some region. On region $(a)$ cylinder $(2)$ is taller than cylinder $(1)$ therefore cylinder $(1)$ acts are bounds for floor and ceiling.

enter image description here

In method $(1)$ you take a circular region, and use another cylinder for bounds. enter image description here

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  • $\begingroup$ thanks to your help I managed to solve it correctly :) $\endgroup$ – Oria Gruber May 11 '14 at 18:07
  • $\begingroup$ @OriaGruber you are welcome :) $\endgroup$ – Santosh Linkha May 11 '14 at 18:08
  • $\begingroup$ What I changed was, basically the boundary for $y$ was incorrect. I changed it to $-\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}$, same as $z$, and $x$ still goes between $-1$ and $1$ and the answer is correct. Now I wonder, the volume of what shape did i calculate before? when I was wrong $\endgroup$ – Oria Gruber May 11 '14 at 18:12
  • $\begingroup$ @OriaGruber I find this little un-intuitive, I prefer method $(2)$ ... let me add a little bit on method $(2)$. $\endgroup$ – Santosh Linkha May 11 '14 at 18:14
  • $\begingroup$ @OriaGruber find the volume of the region in above figure which is triangular and height is curved cylinder and then multiply it by $4$ $\endgroup$ – Santosh Linkha May 11 '14 at 18:19

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