Section 15.2 : Iterated Integrals

1. 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. \[\iint\limits_{R}{{12x - 18y\,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ { - 1,4} \right] \times \left[ {2,3} \right]\] Solution

For problems 2 – 8 compute the given double integral over the indicated rectangle.

1. \( \displaystyle \iint\limits_{R}{{6y\sqrt x - 2{y^3}\,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ {1,4} \right] \times \left[ {0,3} \right]\) Solution
2. \( \displaystyle \iint\limits_{R}{{\frac{{{{\bf{e}}^x}}}{{2y}} - \frac{{4x - 1}}{{{y^2}}}\,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ { - 1,0} \right] \times \left[ {1,2} \right]\) Solution
3. \( \displaystyle \iint\limits_{R}{{\sin \left( {2x} \right) - \frac{1}{{1 + 6y}}\,dA}}\hspace{0.25in}R = \left[ {\frac{\pi }{4},\frac{\pi }{2}} \right] \times \left[ {0,1} \right]\) Solution
4. \( \displaystyle \iint\limits_{R}{{y{{\bf{e}}^{{y^{\,2}} - 4x}}\,dA}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}R = \left[ {0,2} \right] \times \left[ {0,\sqrt 8 } \right]\) Solution
5. \( \displaystyle \iint\limits_{R}{{x{y^2}\,\sqrt {{x^2} + {y^3}} \,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ {0,3} \right] \times \left[ {0,2} \right]\) Solution
6. \( \displaystyle \iint\limits_{R}{{xy\cos \left( {y\,{x^2}} \right)\,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ { - 2,3} \right] \times \left[ { - 1,1} \right]\) Solution
7. \( \displaystyle \iint\limits_{R}{{xy\cos \left( y \right) - {x^2}\,dA}}\hspace{0.25in}\hspace{0.25in}R = \left[ {1,2} \right] \times \left[ {\frac{\pi }{2},\pi } \right]\) Solution
8. Determine the volume that lies under \(f\left( {x,y} \right) = 9{x^2} + 4xy + 4\) and above the rectangle given by \(\left[ { - 1,1} \right] \times \left[ {0,2} \right]\) in the \(xy\)-plane. Solution