# Surface area common to two perpendicularly intersecting cylinders

I need help to calculate the following surface area:

the surface area common to the two cylinders $x^2 + y^2 = a^2$ and $x^2 + z^2 = a^2$ using surface integrals essentially.

My attempt: Let surface area = $S$ and $\hat n = \nabla(x^2 + y^2) = \frac{1}{a}(x \hat i + y \hat j)$ $$S = \int \int ds = \int \int \frac{dy dz}{|\hat n\cdot\hat i|}$$ $$= \int \int \frac{a}{x}dy dz$$ $$= \int \int \frac{a}{\sqrt{a^2 - y^2}}dy dz$$

How do I progress from here on?

• Steinmetz solid – achille hui Oct 13 '15 at 16:26
• @achillehui I saw the result there. But can you show me the way in my attempt? I will be very happy if you help me. I need it very much. – SchrodingersCat Oct 13 '15 at 16:47
• Using symmetry, split the surface into four pieces which give equal contribution to the area. On one of the piece, say the one with $x > 0, |z| \le \sqrt{a^2-y^2}, |y| \le a$. If you integrate the $z$ term first, you will obtain a factor $2\sqrt{a^2-y^2}$ which cancel the corresponding one in denominator. The integral become trivial after that. – achille hui Oct 13 '15 at 17:05
• @achillehui Very very thanks. Can you help me solve another question of mine.... math.stackexchange.com/questions/1476246/… .....I will be eagerly waiting for your reply. Please help. – SchrodingersCat Oct 13 '15 at 17:18