evaluate the double integral $\\iint_d x\\cos y da$, where $d$ is bounded by $y = 0$, $y = x^2$, and $x = 3$.

Explanation:

Step1: Set up the double - integral

Since (D) is bounded by (y = 0,y=x^{2},x = 3), we can set up the double - integral as (\int_{0}^{3}\int_{0}^{x^{2}}x\cos ydydx). First, integrate with respect to (y).

Step2: Integrate with respect to (y)

Using the integral formula (\int\cos ydy=\sin y + C), we have: (\int_{0}^{3}\left[x\sin y\right]{y = 0}^{y=x^{2}}dx=\int{0}^{3}x\sin(x^{2})dx)

Step3: Integrate with respect to (x)

Let (u=x^{2}), then (du = 2xdx) and (xdx=\frac{1}{2}du). When (x = 0), (u = 0); when (x = 3), (u = 9). The integral (\int_{0}^{3}x\sin(x^{2})dx=\frac{1}{2}\int_{0}^{9}\sin udu) Using the integral formula (\int\sin udu=-\cos u + C), we get (\frac{1}{2}[-\cos u]_{0}^{9}=\frac{1}{2}(1-\cos9))

Answer:

(\frac{1}{2}(1 - \cos9))