Integrating circles

Question

Suppose I want to integrate a circle.

$$y^2 + (x-5)^2 = 5^2$$

To find the area of said circle, I'd pick my limits to be $0$ and $5$. However, this circle is on either side of the $x$-axis - what would happen if were to integrate this with respect to $x$? Would the answer be $0$?

Which leads me on to my next question: I know that the volume of a revolution is found by

$$\pi \int y^2 \,\mathrm{d}x.$$

So if I were to then integrate this circle by simply moving the $(x-5)^2$ term to the other side, would I get the volume of two circles?

Answer 1

If you want to find the area of the circle in cartesian coordinates, you can integrate in either $x$ or $y$ and exploit symmetry as follows: $$ A=2\int_0^{10}\sqrt{25-(x-5)^2}\mathrm dx $$ which will require a trig substitution. Similarly, $$ A=2\int_{-5}^5\sqrt{5^2-y^2}+5\mathrm dy $$ which again requires a trig sub.

I don't know what you mean by the volume of two circles. However, if you compute the volume of revolution, you'll get, well, a volume. Specifically the volume of the sphere (if you set up the bounds correctly).

Answer 2

Regarding your first question, $$y^2 + (x-5)^2 = 5^2 \implies y=\pm \sqrt {5^2 -(x-5)^2}$$

The area of the circle is the area bounded by these two semicircles so you need to integrate $$ \int _0^{10} 2\sqrt {5^2 -(x-5)^2}$$ to find the area of the circle.

Regarding your second question you are correct in in finding $y^2$ from your equation.