Electric charge is distributed over the disk $x^2 + y^2 \leq 5$...Find the total charge on the disk.

Question

Electric charge is distributed over the disk $x^2 + y^2 \leq 5$ so that the charge density at $(x,y$) is $\sigma(x,y) = 2 + x^2 + y^2$ coulombs per square meter. Find the total charge on the disk.

$$\int_0^{2\pi}\int_0^5(2+r^2)\space r\space dr\space d\theta=\frac{725\pi}2$$

is not the right answer. I don't know why. Any help?

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The radius is $\sqrt{5}$, not $5$. You have the wrong limit of integration. — IAmNoOne, May 01 '14 at 02:57

score 4 · Accepted Answer · answered May 01 '14 at 02:59

4

The radius of the disk is $\sqrt{5}$, not $5$. Change your upper integration limit and you'll get the right answer. (The formula is $x^2+y^2=r^2$, not $\dots = r$.)

answered May 01 '14 at 02:59

Eric Towers

67,037

$=\frac{45\pi}2$...Thanks – user5826 May 01 '14 at 03:01

Electric charge is distributed over the disk $x^2 + y^2 \leq 5$...Find the total charge on the disk.

1 Answers1