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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 15.2 : Iterated Integrals
- Compute the following double integral over the indicated rectangle (a) by integrating with respect to x first and (b) by integrating with respect to y first. \[\iint\limits_{R}{{16xy - 9{x^2} + 1\,dA}}\hspace{0.5in}R = \left[ {2,3} \right] \times \left[ { - 1,1} \right]\]
- Compute the following double integral over the indicated rectangle (a) by integrating with respect to x first and (b) by integrating with respect to y first. \[\iint\limits_{R}{{\cos \left( x \right)\sin \left( y \right)\,dA}}\hspace{0.5in}R = \left[ {\frac{\pi }{6},\frac{\pi }{4}} \right] \times \left[ {\frac{\pi }{4},\frac{\pi }{3}} \right]\]
For problems 3 – 16 compute the given double integral over the indicated rectangle.
- \( \displaystyle \iint\limits_{R}{{8{x^3} - 4\,dA}}\hspace{0.5in}\hspace{0.25in}R = \left[ { - 3, - 1} \right] \times \left[ {0,4} \right]\)
- \( \displaystyle \iint\limits_{R}{{15{y^4} + \frac{2}{{{x^2}}}\,dA}}\hspace{0.5in}R = \left[ {1,2} \right] \times \left[ {1,4} \right]\)
- \( \displaystyle \iint\limits_{R}{{4y{{\sec }^2}\left( x \right) + \frac{{2x}}{y}\,dA}}\hspace{0.25in}R = \left[ {0,\frac{\pi }{4}} \right] \times \left[ {1,5} \right]\)
- \( \displaystyle \iint\limits_{R}{{{y^5} - {x^2}{{\bf{e}}^y}\,dA}}\hspace{0.5in}\hspace{0.25in}R = \left[ { - 1,2} \right] \times \left[ { - 3,3} \right]\)
- \( \displaystyle \iint\limits_{R}{{\frac{{{x^3}}}{{1 + {x^4}}} - \frac{1}{{{{\bf{e}}^{3y}}}}\,dA}}\hspace{0.5in}R = \left[ { - 1,0} \right] \times \left[ {0,4} \right]\)
- \( \displaystyle \iint\limits_{R}{{x{{\bf{e}}^{{x^{\,2}}}} - 12{x^3}\sin \left( {\pi y} \right)\,dA}}\hspace{0.25in}R = \left[ { - 2,0} \right] \times \left[ {\frac{1}{2},1} \right]\)
- \( \displaystyle \iint\limits_{R}{{x\cos \left( {4y + 3{x^2}} \right)\,dA}}\hspace{0.5in}R = \left[ {0,\sqrt \pi } \right] \times \left[ {\frac{\pi }{2},\pi } \right]\)
- \( \displaystyle \iint\limits_{R}{{\frac{{\ln \left( {4xy} \right)}}{{xy}}\,dA}}\hspace{0.5in}\hspace{0.25in}R = \left[ {1,2} \right] \times \left[ {3,4} \right]\)
- \( \displaystyle \iint\limits_{R}{{{x^2}{y^3}\,{{\bf{e}}^{{x^{\,3}} - {y^{\,4}}}}\,dA}}\hspace{0.5in}R = \left[ {0,1} \right] \times \left[ { - 1,0} \right]\)
- \( \displaystyle \iint\limits_{R}{{42y{x^3}{{\left( {1 + {x^2}{y^2}} \right)}^6}\,dA}}\hspace{0.5in}R = \left[ {0,1} \right] \times \left[ {0,2} \right]\)
- \( \displaystyle \iint\limits_{R}{{\frac{{\cos \left( {\frac{x}{y}} \right)}}{{{y^3}}}\,dA}}\hspace{0.5in}\hspace{0.25in}R = \left[ {\frac{\pi }{2},\pi } \right] \times \left[ {1,2} \right]\)
- \( \displaystyle \iint\limits_{R}{{\frac{{x\cos \left( {\frac{x}{y}} \right)}}{{{y^2}}}\,dA}}\hspace{0.5in}R = \left[ {\frac{\pi }{2},\pi } \right] \times \left[ {1,2} \right]\)
- \( \displaystyle \iint\limits_{R}{{2y\ln \left( x \right) - 20{x^3}{y^3}\,dA}}\hspace{0.25in}R = \left[ {1,2} \right] \times \left[ {0,4} \right]\)
- \( \displaystyle \iint\limits_{R}{{xy{{\bf{e}}^x}\cos \left( y \right)\,dA}}\hspace{0.5in}R = \left[ {0,1} \right] \times \left[ {0,\pi } \right]\)
- Determine the volume that lies under \(f\left( {x,y} \right) = 20 - 3{x^3} - 3{y^2}\) and above the rectangle given by \(\left[ { - 2,2} \right] \times \left[ { - 1,1} \right]\) in the \(xy\)-plane.
- Determine the volume that lies under \(f\left( {x,y} \right) = 10 + xy\sin \left( {{x^2} - {y^2}} \right)\) and above the rectangle given by \(\left[ { - 3,0} \right] \times \left[ {1,3} \right]\) in the \(xy\)-plane.