Answer
$\dfrac{1-\cos (1)}{2}$
Work Step by Step
The domain $D$ in Type -1 can be expressed as follows:
$D=\left\{ (x, y) | 0 \leq y \leq x^2 , \ 0 \leq x \leq 1 \right\}
$
Our aim is to calculate the area of the given surface.
$A=\iint_{D} x \cos y dA \\=\int_{0}^{1} \int_{0}^{x^2} x \cos y \ dy \ dx \\= \int_0^1 x \sin x^2 d x $
Consider $x^2 =u$ and $2x dx = du$
Now, $A=\int_0^1 \dfrac{\sin u du}{2} \\ =\dfrac{1}{2} \int_{0}{1} \sin u du \\ = \dfrac{1} {2}[ -\cos u]_0^1 \\= \dfrac{1-\cos (1)}{2}$
