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I'm solving a list of exercises of double integrals and they normally have a range for $x$ and $y$, but in this case it says that $y = x^2$ and $y = 4$, so I thought that $x$ would be $\sqrt{y}$, but my answer was wrong. The answer should be 25,60).

A thin metal plate occupies a shadow below the figure below.

graph

The region is limited by the graphs of $y = x^2$ and $y = 4$ where x and y are measured in centimeters. If the superficial density of the plate, in $g/cm^2$, is $p(x,y) = y$, its mass, in grams, will be:

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Step 1: You need to translate information "the region is limited" by two curves into inequalities for $x$ and $y$ to describe the shadowed $D$.

For example, $D=\{(x,y)\colon -2\le x\le 2,\,x^2\le y\le 4\}$.

Step 2: After that the problem is to calculate $$ M=\iint_D\rho(x,y)\,dxdy. $$

Iterated integration $$M=\int_{-2}^2\int_{x^2}^4y\,dy\,dx.$$

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Well there are 2 approaches.

Firstly see the integral you want to compute is simply a box minus the area underneath the curve $y=x^2$. That is easily done by finding the intersection points.

OR one may flip the curve $y=x^2$ and then simply find the area under the new curve $y=-x^2+4$ and the $x$-axis.

The mass is then calculated by the answer from A.Γ.

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    $\begingroup$ This is not true. Mass = density $\times$ area works only for constant densities. Here one has to integrate the density. $\endgroup$ – A.Γ. Jun 17 '18 at 1:27
  • $\begingroup$ Ah I completely missed that. $\endgroup$ – Tony Hellmuth Jun 17 '18 at 1:28

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