Calculate $\iint_R \cos\left(\frac{y}{x}\right)\text{d}A$

Question

Hello I would like to know if I answered the question correctly,

Calculate $\iint_R \cos\left(\frac{y}{x}\right)\text{d}A$ with $R$ the region under the graph of $y = x^3$, between $x = 0$ and $x = 1$.

I calculated the following integral with substitution at first with $u = g(y)= \frac{y}{x} = \frac{1}{x}y$, that gives $\frac{\text{d}u}{\text{d}y} = \frac{1}{x}$. In the sixth equation I used partial integration $f(x)g'(x)\text{d}x = f(x)g(x) - \int g(x)f'(x) \text{d}x$ with $f(x) = x, f'(x) = 1, g'(x) = \sin(x)$ and $g(x) = -\cos(x)$. \begin{align*} \int_0^1\int_0^{x^3} \cos \left(\frac{y}{x}\right) \text{d}y\text{d}x &= \int_0^1\int_{g(0)}^{g(x^3)} \cos(u) x\text{d}u\text{d}x \\ &=\int_0^1x\int_{\frac{0}{x}}^{\frac{x^3}{x}} \cos(u) \text{d}u\text{d}x\\ &= \int_0^1x\cdot \left[\sin(u)\right]_0^{x^2} \text{d}x\\ &= \int_0^1x\cdot \left[\sin(\frac{y}{x})\right]_0^{x^2} \text{d}x\\ &= \int_0^1 x \sin(x) \text{d}x \\ & = \left[x(-\cos(x)) - \int -\cos(x) \text{d}x \right]_0^1 \\ &= \left[-x \cos(x) +\sin(x)\right]_0^1 \\ &= -1\cos(1)+\sin(1) - \left(-0 \cos(0) + \sin(0)\right) \\ &= -\cos(1) + \sin(1). \end{align*}

Answer 1

There's no need to use a substitution and worry about new bounds of integration: $$\int_0^1\int_0^{x^3} \cos \left(\frac{y}{x}\right) \text{d}y\text{d}x =\int_0^1\int_0^{x^3}\frac{x}{x} \cos \left(\frac{y}{x}\right) \text{d}y\text{d}x=\int_0^1 x\left[\sin \left(\frac{y}{x}\right)\right]_{y=0}^{y=x^3}\text{d}x= \\ =\int_0^1 x \sin (x^2)\text{d}x=-\frac{1}{2}\cos x^2 \bigg \vert_0^1 =\frac{1}{2}\left(1-\cos 1 \right) $$

Answer 2

Yes, your calculation is mostly correct, but after the first substitution, you should have $x^2$ as upper bound and not $x^3$.

However, you should maybe state more clearly that $R$ is the area bounded by the graphs $y=x^3$ and $y=0$ in between $x=0$ and $x=1$. Else, it is not clear what the lower $y$-bound of the domain is.