Question #128002

determine the mass of the lamina corresponding to the first quadrant portion of the circle x^2+y^2 =25 where the density at thr point of (x,y) is f(x,y)=k*sqrt(x^2+y^2)

Expert's answer

∬x2+y2≤25, x≥0, y≥0kx2+y2 dx dy=\iint_{x^2 + y^2 \leq 25, \ x \geq 0, \ y \geq 0} k \sqrt{x^2 + y^2} \,dx\,dy =






=∫05 dx∫025−x2kx2+y2 dy== \int_{0}^{5} \,dx \int_{0}^{\sqrt{25 -x^2}} k \sqrt{x^2 + y^2} \,dy =


{x=rcosϕy=rsinϕ\begin{cases} x = rcos\phi \\ y = rsin\phi\\ \end{cases} \\

=∫0π/2 dϕ∫05k(rcosϕ)2+(rsinϕ)2 rdr== \int_{0}^{\pi/2} \,d\phi \int_{0}^{5} k \sqrt{(rcos\phi)^2 + (rsin\phi)^2} \,rdr =

=∫0π/2 dϕ∫05kr2 dr=∫0π/2k∗53/3 dϕ=k∗53/3∗π/2=125πk/6= \int_{0}^{\pi/2} \,d\phi \int_{0}^{5} kr^2 \,dr = \int_{0}^{\pi/2} k *5^3/3 \,d\phi = k *5^3/3 * \pi/2 = 125\pi k/6




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