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I just learned that the single variable definitely integral is the area under the curve via the Riemann's sum intuitive, but now I'm lost as to why the double integral also represents an area? What is going on?

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Like in the last example, isn't the inner integral the area of the shaded region from $X^{2}$ to 4

Jwan622
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    The double integral is only equal to the area when $f(x)=1$ – tomi Oct 06 '20 at 13:16
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    You could think of a double integral as computing the volume under the graph of $f$. But if $f(x,y) = 1$ for all $x,y$ then the volume of that region is area of base $\times$ height = area of base $\times 1$ = area of the region you're integrating over. – littleO Oct 06 '20 at 13:19
  • Another viewpoint: if $E \subset \mathbb R$ is an interval, then $\int_E 1 , dx$ is equal to the length of $E$. Your question is kind of like asking "why does a single integral tell us the length of an interval". The analogous fact for double integrals is that $\int \int_R 1 , dx , dy$ is equal to the area of $R$. – littleO Oct 06 '20 at 13:22
  • So if the double integral is only equal to the area when f(x) = 1, what is my textbook saying in re to region R? – Jwan622 Oct 06 '20 at 16:29
  • @Jwan622 It's just saying that $\int \int_R 1 , dx , dy = \text{area of } R$. In the equation after the sentence you highlighted, the integrand is $1$. – littleO Oct 06 '20 at 16:53
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    @tomi Though you are (pretty much) correct, it would have been less confusing to the uninitiated to say "when $f(x,y)=1$". – Servaes Oct 06 '20 at 20:00
  • @Servaes Yes, my bad. Shame I can't edit that comment... – tomi Oct 07 '20 at 14:54

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Just like in general the integral $\int_If(x)\,\mathrm{d}x$ measures the area under the graph of $f$ in the interval $I$, so too does the double integral $\iint_Rg(x,y)\,\mathrm{d}x\mathrm{d}y$ measure the volume under the graph of $g$ in the region $R$. However, for the constant function $f(x)=1$ the integral $$\int_If(x)\,\mathrm{d}x=\int_I\,\mathrm{d}x,$$ measures the area of a rectangle of sides $1$ and $I$, so effectively it measures the length of $I$. Similarly for the constant function $g(x,y)=1$ the integral $$\iint_Rg(x,y)\,\mathrm{d}x\mathrm{d}y=\iint_R\,\mathrm{d}x\mathrm{d}y,$$ effectively measures the area of $R$.

Servaes
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  • So if the double integral is only equal to the area when f(x) = 1, what is my textbook saying in re to region R? – Jwan622 Oct 06 '20 at 16:29
  • The text explicitly states (which you have highlighted) that the area of a region $R$ is equal to the specific double integral $\iint_R,\mathrm{d}x\mathrm{d}y$. – Servaes Oct 06 '20 at 18:50
  • Why does someone comment: "The double integral is only equal to the area when ()=1 " – Jwan622 Oct 06 '20 at 19:29