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Part 1Evaluate the following integral in cylindrical coordinates.ModifyingAbove ModifyingBelow Integral from nothing to nothing With 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With 3 StartRoot 2 EndRoot divided by 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 9 minus x squared EndRoot width x e Superscript negative x squared minus y squared Baseline dy font size decreased by 4 dx font size decreased by 4 dz2∫ 032/2∫0 9−x2∫xe−x2−y2dy dx dz Part 1ModifyingAbove ModifyingBelow Integral from nothing to nothing With 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With 3 StartRoot 2 EndRoot divided by 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 9 minus x squared EndRoot width x e Superscript negative x squared minus y squared Baseline dy font size decreased by 4 dx font size decreased by 4 dz2∫ 032/2∫0 9−x2∫xe−x2−y2dy dx dzequals=[input]enter your response here ​(Type an exact​ answer, using piπ as​ needed.)

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The prompt presents a multipart integral to evaluate in cylindrical (polar) coordinates, with the integrand involving sqrt(9 - x^2) times x times e^{-(x^2 + y^2)}, and an accompanying z-variable with its own limits. Since the formatting is unclear, I will focus on the provided answer option and how such a result could arise, while outlining what each part of the integral typically contributes. Option being analyzed: -\frac{3\sqrt{2}\pi}{2}\,(1 - e^{9})\,e^{-9} First, note the algebraic simplification of this expression. The factor (1 - e^{9})e^{-9} can be rewritten via distributive properties as e^{-9} - 1. Multiplying by the negative sign in front yields (3√2π/2) (1 - e^{-9}). Recognizing this form helps us see a common pattern: results of Gaussian-type integrals over......Login to view full explanation

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Part 1Evaluate the following integral in cylindrical coordinates.ModifyingAbove ModifyingBelow Integral from nothing to nothing With 5 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 25 minus x squared EndRoot width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot x squared plus y squared EndRoot width 0 left parenthesis x squared plus y squared right parenthesis Superscript negative 1 divided by 2 Baseline dz font size decreased by 4 dy font size decreased by 4 dx5∫ 025−x2∫0 x2+y2∫0x2+y2−1/2 dz dy dx Part 1ModifyingAbove ModifyingBelow Integral from nothing to nothing With 5 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 25 minus x squared EndRoot width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot x squared plus y squared EndRoot width 0 left parenthesis x squared plus y squared right parenthesis Superscript negative 1 divided by 2 Baseline dz font size decreased by 4 dy font size decreased by 4 dx5∫ 025−x2∫0 x2+y2∫0x2+y2−1/2 dz dy dxequals=[input]enter your response here ​(Type an exact​ answer, using piπ as​ needed.)

Part 1Evaluate the following integral in cylindrical coordinates.ModifyingAbove ModifyingBelow Integral from nothing to nothing With 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With 3 StartRoot 2 EndRoot divided by 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 9 minus x squared EndRoot width x e Superscript negative x squared minus y squared Baseline dy font size decreased by 4 dx font size decreased by 4 dz2∫ 032/2∫0 9−x2∫xe−x2−y2dy dx dz Part 1ModifyingAbove ModifyingBelow Integral from nothing to nothing With 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With 3 StartRoot 2 EndRoot divided by 2 width 0 ModifyingAbove ModifyingBelow Integral from nothing to nothing With StartRoot 9 minus x squared EndRoot width x e Superscript negative x squared minus y squared Baseline dy font size decreased by 4 dx font size decreased by 4 dz2∫ 032/2∫0 9−x2∫xe−x2−y2dy dx dzequals=[input]negative \frac StartSet 3−\frac{3 ​(Type an exact​ answer, using piπ as​ needed.)

Part 1[table] Use a triple integral to find the volume of the solid bounded by the surfaces zequals=2e Superscript yey and zequals=22 over the rectangle StartSet left parenthesis x,y right parenthesis : 0 less than or equals x less than or equals 1, 0 less than or equals y less than or equals ln 4 EndSet{(x,y): 0≤x≤1, 0≤y≤ln4}. | ln 4ln411xxyyzz [/table] Part 1The volume of the solid is [input]enter your response here [input] ▼  units. cubic units. square units. empty selection ​(Type an exact​ answer.)

Part 1Evaluate the following integral.Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dV​; Dequals=​{(x,y,z): minus−33less than or equals≤xless than or equals≤33​, minus−44less than or equals≤yless than or equals≤44​, minus−11less than or equals≤zless than or equals≤11​} Part 1Integral from nothing to nothing ModifyingBelow Integral from nothing to nothing With Upper D Integral from nothing to nothing left parenthesis xy plus xz plus yz right parenthesis dV∫∫D∫(xy+xz+yz) dVequals=[input]enter your response here ​(Simplify your​ answer.)

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