Question

Find the total charge on the given disk. The disk is\({x^2} + {y^2} \le 1\). The charge density is\(\sigma (x,y) = \sqrt {{x^2} + {y^2}} .\)

Short answer

The total charge on the given rectangle is \(\frac{{2\pi }}{3}{\rm{C}}.\)

Step-by-step solution

Formula of the total charge\(Q\)and the concept of polar region

The total charge\(Q\)on the given rectangular region\(R\)is,

Here, the charge density is given by\(\sigma (x,y)\)

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta {\rm{,}}\)where\(0 \le \beta - \alpha \le 2\pi {\rm{,}}\)then,

If\(g(x)\)is the function of\(x\)and\(h(y)\)is the function of\(y\)then,

\(\int_a^b {\int_c^d g } (x)h(y)dydx = \int_a^b g (x)dx\int_c^d h (y)dy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,........\left( 2 \right)\)

Use the formula of total charge and the concept of polar region to find the total charge on the given disk

The total charge \(Q\)on the given rectangle \(R\)is,

Substitute \(x = r\cos \theta ,y = \sin \theta \) and convert the coordinates into polar coordinates.

Therefore, by equation (1), the integral becomes,

\(\begin{aligned}{c}Q = \int_0^1 {\int_0^{2\pi } {\sqrt {{r^2}} } } (r)d\theta dr\\ = \int_0^1 {\int_0^{2\pi } r } (r)d\theta dr\\ = \int_0^1 {\int_0^{2\pi } {{r^2}} } d\theta dr\end{aligned}\)

Integrate with respect to \(r\)and \(\theta \) by using (2),

\(\begin{aligned}{c}Q = \int_0^1 {{r^2}} dr\int_0^{2\pi } d \theta \\ = \left( {\frac{{{r^3}}}{3}} \right)_0^1(\theta )_0^{2\pi }\end{aligned}\)

Apply the limit value for\(r\)and\(\theta \),

\(\begin{aligned}{c}Q = \left( {\frac{{{1^3}}}{3}} \right.\left. { - \frac{{{0^3}}}{3}} \right)(2\pi - 0)\\ = \frac{1}{3}(2\pi )\\ = \frac{{2\pi }}{3}\end{aligned}\)

Thus, the total charge on the given rectangle is \(\frac{{2\pi }}{3}\,{\rm{C}}.\)