Section 15.3 : Double Integrals over General Regions
- Evaluate \( \displaystyle \iint\limits_{D}{{8y{x^3}\,dA}}\) where \(D = \left\{ {\left( {x,y} \right)| - 1 \le y \le 2, - 1 \le x \le 1 + {y^2}} \right\}\)
- Evaluate \( \displaystyle \iint\limits_{D}{{12{x^2}y - {y^2}\,dA}}\) where \(D = \left\{ {\left( {x,y} \right)| - 2 \le x \le 2, - {x^2} \le y \le {x^2}} \right\}\)
- Evaluate \( \displaystyle \iint\limits_{D}{{9 - \frac{{6{y^2}}}{{{x^2}}}\,dA}}\) where \(D\) is the region in the 1st quadrant bounded by \(y = {x^3}\) and \(y = 4x\).
- Evaluate \( \displaystyle \iint\limits_{D}{{15{x^2} - 6y\,dA}}\) where \(D\) is the region bounded by \(x = \frac{1}{2}{y^2}\) and \(x = 4\sqrt y \).
- Evaluate \( \displaystyle \iint\limits_{D}{{6y{{\left( {x + 6} \right)}^2}\,dA}}\) where \(D\) is the region bounded by \(x = - {y^2}\) and \(x = y - 6\).
- Evaluate \( \displaystyle \iint\limits_{D}{{{{\bf{e}}^{{y^{\,2}} + 1}}\,dA}}\) where \(D\) is the triangle with vertices \(\left( {0,0} \right)\), \(\left( { - 2,4} \right)\) and \(\left( {8,4} \right)\).
- Evaluate \( \displaystyle \iint\limits_{D}{{7{y^3}{{\bf{e}}^{{x^{\,2}} + 1}}\,dA}}\) where \(D\) is the region bounded by\(y = 2\,\,\sqrt[4]{x}\), \(x = 9\) and the \(x\)-axis.
- Evaluate \( \displaystyle \iint\limits_{D}{{{x^5}\sin \left( {{y^4}} \right)\,dA}}\) where \(D\) is the region in the 2nd quadrant bounded by \(y = 3{x^2}\), \(y = 12\) and the \(y\)-axis.
- Evaluate \( \displaystyle \iint\limits_{D}{{xy - {y^2}\,dA}}\) where \(D\) is the region shown below.

- Evaluate \( \displaystyle \iint\limits_{D}{{12{x^3} - 3\,dA}}\) where \(D\) is the region shown below.

- Evaluate \( \displaystyle \iint\limits_{D}{{6{y^2} + 10y{x^4}\,dA}}\) where \(D\) is the region shown below.

- Evaluate \( \displaystyle \iint\limits_{D}{{\frac{{{x^3}}}{{{y^2}}}\,dA}}\) where \(D\) is the region bounded by \(\displaystyle y = \frac{1}{{{x^2}}}\), \(x = 1\) and \(\displaystyle y = \frac{1}{4}\) in the order given below.
- Integrate with respect to \(x\) first and then \(y\).
- Integrate with respect to \(y\) first and then \(x\).
- Evaluate \( \displaystyle \iint\limits_{D}{{xy - {y^3}\,dA}}\) where \(D\) is the region bounded by \(y = {x^2}\), \(y = - {x^2}\) and \(x = 2\) in the order given below.
- Integrate with respect to \(x\) first and then \(y\).
- Integrate with respect to \(y\) first and then \(x\).
For problems 14 – 16 evaluate the given integral by first reversing the order of integration.
- \(\displaystyle \int_{0}^{8}{{\int_{{{y^{\frac{1}{3}}}}}^{2}{{\frac{y}{{{x^7} + 1}}\,dx}}\,dy}}\)
- \(\displaystyle \int_{{ - 4}}^{0}{{\int_{{\sqrt { - x} }}^{2}{{{x^{ - \,\,\frac{2}{3}}}\,\,\sqrt {{y^{\frac{5}{3}}} + 1} \,dy}}\,dx}}\)
- \(\displaystyle \int_{0}^{2}{{\int_{{ - x}}^{{3x}}{{5{y^2}{x^3} + 2\,dy}}\,dx}}\)
- Use a double integral to determine the area of the region bounded by \(x = - {y^2}\) and \(x = y - 6\).
- Use a double integral to determine the area of the region bounded by \(y = {x^2} + 1\) and \(\displaystyle y = \frac{1}{2}{x^2} + 3\).
- Use a double integral to determine the volume of the region that is between the \(xy\)‑plane and \(f\left( {x,y} \right) = 2 - x{y^2}\) and is above the region in the \(xy\)-plane that is bounded by \(y = {x^2}\) and \(x = 1\).
- Use a double integral to determine the volume of the region that is between the \(xy\)‑plane and \(f\left( {x,y} \right) = 1 + {y^5}\,\,\sqrt {{x^4} + 1} \) and is above the region in the \(xy\)-plane that is bounded by \(y = \sqrt x \), \(x = 2\) and the \(x\)-axis.
- Use a double integral to determine the volume of the region in the first octant that is below the plane given by \(2x + 6y + 4z = 8\).
- Use a double integral to determine the volume of the region bounded by \(z = 3 - 2y\), the surface \(y = 1 - {x^2}\) and the planes \(y = 0\) and \(z = 0\).
- Use a double integral to determine the volume of the region bounded by the planes \(z = 4 - 2x - 2y\), \(y = 2x\), \(y = 0\) and \(z = 0\).
- Use a double integral to determine the formula for the area of a right triangle with base, \(b\) and height \(h\).
- Use a double integral to determine a formula for the figure below.
