answer:We want to compute the integral {eq}, displaystyle iint_D xy,dA, ; D={(x,y)inmathbb R^2: 1leqslant x^2+y^2 leqslant 4}, {/eq} by using polar coordinates. The transformation equations from Cartesian {eq}, (x,y), {/eq} to Polar {eq}, (r,theta), {/eq} coordinates are {eq}begin{cases} x= rcostheta y = rsintheta end{cases} ; Rightarrow ; xy =r^2sinthetacostheta {/eq}. The Jacobian determinant for that transformation is {eq}, J = r, {/eq}. Then, the transition from an integral in Cartesian coordinates to the integral in Polar coordinates could be represented as: {eq}displaystyle iint_{D_{xy}} f(x,y),dA = displaystyle iint_{D_{rtheta}} f(rcostheta,rsintheta)cdot r, dr,dtheta {/eq}. The region {eq}, D, {/eq} given on the exercise represents a ring determined by two circumferences with radius {eq}, 1 ,text{ and }, 2, {/eq}. Then, the region could be described in Polar coordinates as {eq}, D={(r,theta):,1leqslant rleqslant 2,, 0leqslant thetaleqslant 2pi,} {/eq}. Thus, the integral will be {eq}begin{array}{rl} displaystyle iint_D xy,dA& = displaystyleiint _{D_{rtheta}} r^2sinthetacosthetacdot r,dr,dtheta & = displaystyleint _0^{2pi}sinthetacostheta,dtheta displaystyleint _1^2 r^3,dr & = displaystyleint _0^{2pi}sinthetacostheta,dtheta left(displaystylefrac {r^4}{4}Big|_ {1}^2 right) & = displaystylefrac {15}{4} displaystyleint _0^{2pi}sinthetacostheta,dtheta & = displaystylefrac {15}{4} cdot 0 & = 0 end{array} {/eq}

answer:The simplified form is frac{296196766695424 left(-cos left(frac{2 pi }{9}right)+i sin left(frac{2 pi }{9}right)right)}{9765625}, which can be expressed in trigonometric form as 296196766695424 cdot left(-cos left(frac{2 pi }{9}right)right) + i cdot 296196766695424 cdot sin left(frac{2 pi }{9}right). Note: The answer has been checked for accuracy, and the trigonometric identity cos(10A) - isin(10A) = (cos A + isin A)^{10} has been correctly applied to obtain the result.

answer:The integral of {eq}xcos(3x) , dx{/eq} can be found using the substitution method. Let's start by rewriting the integral: {eq}int xcos(3x) , dx = frac{1}{3}int (3x)cos(3x) , dx{/eq} Now, make the substitution: {eq}u = 3x Rightarrow frac{du}{dx} = 3 Rightarrow dx = frac{du}{3}{/eq} Substituting this into the integral: {eq}begin{align*} int xcos(3x) , dx &= frac{1}{3}int ucos(u)left(frac{du}{3}right) &= frac{1}{9}int ucos(u) , du end{align*}{/eq} To evaluate the integral, we'll use the product rule in reverse (integration by parts): {eq}begin{align*} frac{1}{9}int ucos(u) , du &= frac{1}{9}left[usin(u) - int sin(u) , duright] &= frac{1}{9}left[usin(u) + cos(u)right] &= frac{1}{9}(3xsin(3x) + cos(3x)) &quad text{(Replace } u text{ with } 3x) end{align*}{/eq} Therefore, the integral of {eq}xcos(3x) , dx{/eq} is: {eq}frac{1}{9}(3xsin(3x) + cos(3x)){/eq}

answer:Prevailing interest rates in the two currencies involved in the transaction.