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.