How to prove this function to be identically zero I need help in this particular question due to the reason that I am not very comfortable in questions when integrals are to be proved identically zero.

Question Let $f:[0,1]\times[0,1]\longrightarrow[0,\infty)$ be a continuous function . Suppose that $$\int_{0}^{1}\bigg(\int_{0}^{1} f(x,y)dy\bigg)dx=0.$$ Then prove that $f$ is an identically zero function.

I think this would be due to the fact that range of $f$ is non-negative. But In such kind of questions I am not able to rigorously prove these. Actually, I was taught analysis by a really poor instructor and problem solving/assignments in our university are minimal. So, I use to try exercises by myself and ask questions here.
Can anyone please tell on how should I attempt this particular problem.
I shall be really thankful.
Hint :

Prove the result in the one-dimensional case : that is, if $\varphi : [0,1] \rightarrow \mathbb{R}$ is a non-negative and continuous function such that
$$\int_0^1 \varphi(x) \mathrm{dx} = 0$$
then $\varphi = 0$ (maybe you already know how to prove that).

Apply this result to the function $x \mapsto \int_0^1 f(x,y) \mathrm{dy}$ (check all the hypothesis correctly !) to deduce that for all $x \in [0,1]$, you have $\int_0^1 f(x,y) \mathrm{dy} = 0$.

Re-apply the result to the function $y \mapsto f(x,y)$ with $x$ fixed to deduce finally that for all $x,y$, you have $f(x,y)=0$.
Since the function is strictly non-negative we can interpret this integral as the volume under the graph of the function on the square $[0,1]\times [0,1]$. Now assume there exists some point $(x_0,y_0)$ such that $f(x_0,y_0) \neq 0$. Since the function is continuous there exists some small neighborhood of $(x_0,y_0)$ where all $f(x,y) &gt; 0$ in that neighborhood, by the intermediate value theorem. Since this neighborhood is strictly positive, it must have some positive, non-zero volume. This is impossible because the volume is given as zero.