Computing area of a set

Question

Problem

Define area of set $D$ when set $D$ is defined as:

$$ D=\{(x,y)| -1 \le x \le 1, x^2 \le y \le x^2+2 \} $$

Attempt to solve

Area in terms of $y$-axis bounded between two parabolas: $y=x^2,y=x^2+2$ The latter one is above and first one is below. Our area is bounded in terms of $x$-axis with lines $x=-1,x=1$ which are perpendicular to $x$-axis.

$$A=\int_{-1}^{1} \int_{x^2}^{x^2+2}dydx$$ $$ A=\int_{-1}^{1}(x^2+2-x^2)dx $$ $$ A=\int_{-1}^{1}2dx $$ $$ A=2 \cdot (1) - 2\cdot(-1) $$ $$ A=4 $$

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Now i wish to receive some feedback if this seems correct or if you see flaw in this. That would be highly appreciated.

Answer 1

This is actually correct.

Another way to see this is by interpreting the domain of integration visually.

You are interested in the region below the curve $f = x^2 + 2$, but above the curve $g = x^2$, where $-1 \le x \le 1$.

A simple way to compute this would be: $$\int_{-1}^{1} f(x) dx - \int_{-1}^{1} g(x) dx,$$ which is precisely what you arrived at using double integration.

Be careful, though. This method worked in this case because $f > g$ throughout the region of interest. In general, you can't always make this simplification. However, it is often helpful to draw a sketch to get a better sense of what's going on.

Answer 2

Maybe the source of the problem wanted you to go the way you propose. But be aware that intuitively it is absolutely obvious that the area in question is $4$: Partition the shape into narrow vertical strips. They are all of height $\approx2$, and their total width is $2$.