Calculus III - Double Integrals over General Regions (Assignment Problems)

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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 1^st 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 2^nd 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.
$This region consists of two sub regions. The first sub region is the region in the fourth quadrant bounded by $y=x-1$, the x-axis and the y-axis. The second sub region is the region in the second quadrant bounded by $y=1-x^{2}$, the x-axis and the y-axis.$
Evaluate $ \displaystyle \iint\limits_{D}{{6{y^2} + 10y{x^4}\,dA}}$ where $D$ is the region shown below.
$This region consists of two sub regions. The first sub region is the region in the first and fourth quadrants bounded by $x=2-y^{2}$ and the y-axis. The second sub region is the region in the second and third quadrants bounded by $x=y^{2}-1$ and the y-axis.$
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.
1. Integrate with respect to $x$ first and then $y$.
2. 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.
1. Integrate with respect to $x$ first and then $y$.
2. 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.