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.



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.

  • $\begingroup$ That makes sense. I suppose I just find the double integral to be odd in general. $\endgroup$ – Art Oct 23 '18 at 5:45

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