# RegionPlot of a Double integration [closed]

Hey guys I have a problem where I want to sketch the region of integration:

Integrate[x Cos[y^2], {x, 0, 3}, {y, x^2, 9}]


I know I have to use RegionPlot but it is not working out for me. This is what I am doing: RegionPlot[x^2, {x, 0, 3}, {y, 0, 9}]

Can anyone help me out?

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## closed as off-topic by ubpdqn, Sjoerd C. de Vries, Artes, m_goldberg, Michael E2Apr 13 '14 at 15:40

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Change x^2 to x^2<=y<=9 in the first argument of RegionPlot. –  kguler Apr 13 '14 at 6:09
Thank you! I got it to work but I couldn't answer my own question because I'm a noob. RegionPlot[0 <= x <= 3 && x^2 <= y <= 9, {x, 0, 5}, {y, 0, 10}] –  Tags Apr 13 '14 at 6:13

For illustrative purposes: The boundary of region can be plotted (-3 just to show projection):

Plot3D[{x Cos[y^2], -3}, {x, 0, 3}, {y, 0, 9},
MeshFunctions -> (#2 - #1^2 &), Mesh -> {{0}},
MeshStyle -> {Thick, Red}, PlotPoints -> 50]


The integral is approximately -0.157 (within a region of 'volume'of around 81). This can be visualized by adding plane at z=0:

Plot3D[{x Cos[y^2], 0}, {x, 0, 3}, {y, 0, 9},
RegionFunction -> Function[{x, y}, x^2 <= y <= 9], Mesh -> None,
PlotPoints -> 50, PlotStyle -> {Opacity[0.2], Automatic},
PlotRange -> All]


The region of interest can be plotted:

Plot3D[{x Cos[y^2], -3}, {x, 0, 3}, {y, 0, 9},
RegionFunction -> Function[{x, y}, x^2 <= y <= 9], Mesh -> None,
PlotPoints -> 100]


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