# Intuition on Double Integrals

Frequently, I am met with problem that ask to evaluate a double integral over a bounded region.

For example, evaluate the double integral $$\int\int_R 2x\cos(y)+3 \space dA$$

over the region $$R$$ bounded by $$y=2x^2$$, $$y=0$$, and $$x=1$$. Graphically,

Setting up the integral, I get $$\int_0^1\int_0^{2x^2} 2x\cos(y) \space dy \space dx$$

My question is how exactly does this double integral evaluate the volume above the region R, since the limits of integration seem to have no dependence on the z-axis; the limits of integration are in terms of $$x$$ and $$y$$.

Evaluating the inner integral, wouldn't you get the area between $$y=0$$ and $$y=2x^2$$ of $$2x\cos(y)$$. If so, how does integrating that area between $$x=0$$ and $$x=1$$ give you the volume underneath the surface? The double integral seems random.

The question may be broad, but any intuition on this would be helpful.

Thanks.

But the $$z$$ coordinate appears there, implicitely: $$z=2x\cos(y)$$.
If you compute an integral $$\int_a^bf(x)\,\mathrm dx$$ of a non-negative function $$f$$, that integral is the area of the region below the graph of $$f$$ and above the $$x$$-axis. Similarly, if you compute an integral $$\iint_Af(x,y)\,\mathrm dx\,\mathrm dy$$, that inegral is the volume of the region below the graph of $$f$$ and above the $$xy$$-plane.