Question

Find the mass of the metal disc bounded by the circle x^2+y^2=25 given that the density p(x,y)=e^(sqrt(x^2+y^2))

Answer 1

to compute the mass we merely integrate the planar density on the given region:$$A=\iint\limits_D e^{\sqrt{x^2+y^2}}\,dx\,dy$$where \(D\) is said disk i.e. \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le25\}\)

Answer 2

now recognize that since our region of integration is circular and our integrand has \(x^2+y^2\) it may benefit us to switch to polar coordinates, in which case we let \(\theta\) vary \(0\to2\pi\) and \(r\) \(0\to5\). use the fact \(r^2=x^2+y^2\) to simplify our integrand and recall our new differential after the change of variables is given \(dx\,dy=r\,dr\,d\theta\):$$A=\int_0^{2\pi}\int_0^5 re^r\,dr\,d\theta=2\pi\int_0^5 r e^r\,dr$$

Answer 3

to evaluate this last integral let's use integration by parts to figure out an antiderivative:$$\large\int\underbrace{r}_u\underbrace{e^r\,dr}_{dv}=\underbrace{r}_u\underbrace{e^r}_v-\int\underbrace{e^r}_{v}\,\underbrace{dr}_{du}=re^r-e^r+C$$

Answer 4

so now back to evaluating the mass:$$A=2\pi\int_0^5 re^r\,dr=2\pi\left[(r-1)e^r\right]_0^5=2\pi[4e^5-(-1)e^0]=2\pi(4e^5+1)$$