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1. Use a double integral to find the area of the region that lies below the parabola y = 4x -...

Question:

1. Use a double integral to find the area of the region that lies below the parabola {eq}y = 4x - x^2 {/eq} and above the line 3x + y = 6.

2. Find the volume under the paraboloid {eq}z = 1 - 3x^2 - 3y^2 {/eq} and above the xy-plane.

3. Find the volume of the region bounded above by the paraboloid {eq}z = 1 - x^2 - y^2 {/eq} and below by the plane z = 1 - y.

Volumes of Solids:

The volume of any solid {eq}E {/eq} in {eq}\mathbb{R}^3 {/eq} can be calculated using the triple integral:

$$Volume = \iiint \limits_E dV $$

However, suppose the solid {eq}E {/eq} is described as being the solid between two smooth surfaces {eq}z = F(x,y) \ge G(x,y) {/eq} over the planar region {eq}D {/eq} on the xy-plane.

Then, the volume of the solid can be expressed using only double integrals such that:

$$Volume = \iint \limits_D \left[ F(x,y) - G(x,y) \right] \,dA $$

Answer and Explanation:

1. Let {eq}R {/eq} be the given region which lies below the parabola {eq}y = 4x - x^2 {/eq} and above the line {eq}3x + y = 6 {/eq}.

Graph of Region R

Graph of Region R

Determine the x-coordinates of the intersection points of the parabola and the line:

{eq}\begin{align*} &y = 4x - x^2 = 6 - 3x & \text{[Solve for } x ] \\ &\Rightarrow x^2 - 7x + 6 = 0 \\ &\Rightarrow (x - 1) (x - 6) = 0 \\ &\Rightarrow x - 1 = 0 \text{ or } x - 6 = 0 \\ &\Rightarrow x = 1 \text{ or } x = 6 \end{align*} {/eq}


Based on the graph, the given region {eq}R {/eq} can be described using the inequalities {eq}1 \le x \le 6 \text{ and } 6 - 3x \le y \le 4x - x^2 {/eq}. Then, the area of the region {eq}R {/eq} is given by the double integrals:

{eq}\begin{align*} Area &= \iint \limits_R dA & \text{[Evaluate as iterated integrals]} \\ &= \int_1^6 \int_{6 - 3x}^{4x - x^2} dy \,dx & \text{[Integrate with respect to } y \text{]} \\ &= \int_1^6 \left. y \right|_{6 - 3x}^{4x - x^2} \,dx & \text{[Fundamental Theorem of Calculus]} \\ &= \int_1^6 \left[ ( 4x - x^2 ) - ( 6 - 3x ) \right] \,dx \\ &= \int_1^6 ( -x^2 + 7x - 6 ) \,dx & \text{[Integrate with respect to } x \text{]} \\ &= \left. \left( - \frac{x^3}{3} + \frac{7x^2}{2} - 6x \right) \right|_1^6 & \text{[Fundamental Theorem of Calculus]} \\ &= \left[ - \frac{6^3}{3} + \frac{7 (6)^2}{2} - 6(6) \right] - \left[ - \frac{1^3}{3} + \frac{7 (1)^2}{2} - 6(1) \right] & \text{[Simplify]} \\ &= ( -72 + 126 - 36 ) - \left( - \frac{1}{3} + \frac{7}{2} - 6 \right) \\ &= 18 + \frac{17}{6} \\ Area &= \boxed{ \frac{125}{6} } & \boxed{ \text{Area of the Bounded Region } R } \end{align*} {/eq}


Note that the integral above is similar to the double integrals for the volume of a solid, only that the integrand above is equal to {eq}1 {/eq}.

This is because if the solid is a prism that is bounded above by the plane {eq}z = 1 {/eq} and below by the region {eq}R {/eq}, the solid will have {eq}R {/eq} as its base and height equal to {eq}h =1 {/eq}. Then, the volume of such solid is equal to:

{eq}\begin{align*} &Volume = \iint \limits_R 1 \,dA = Area_{base} \times h \\ &Volume = \iint \limits_R dA = Area_R \end{align*} {/eq}


2. Volume under the paraboloid {eq}z = 1 - 3x^2 - 3y^2 {/eq} and above the xy-plane. Let {eq}D {/eq} be the planar region on the xy-plane that is under the given solid.

Graph of the Solid under the Paraboloid

Graph of the Solid under the Paraboloid

Determine the equation of the intersection of the paraboloid and the xy-plane, which encloses the region {eq}D {/eq}.

{eq}\begin{align*} &z = 1 - 3x^2 - 3y^2 = 0 \\ &\Rightarrow 1 - 3x^2 - 3y^2 = 0 \\ &\Rightarrow 3x^2 + 3y^2 = 1 \\ &\Rightarrow x^2 + y^2 = \frac{1}{3} & \text{[Equation of a circle. Convert into its polar form using } x^2 + y^2 = r^2 ] \\ &\Rightarrow r^2 = \frac{1}{3} & \text{[Take the positive root of } r ] \\ &\Rightarrow r = \frac{1}{\sqrt{3}} \end{align*} {/eq}


Since the region {eq}D {/eq} is circular, it is better to define it in polar coordinates such that {eq}D = \left\{\, (r,\theta) \,|\, 0 \le r \le \frac{1}{\sqrt{3}} \text{ and } 0 \le \theta \le 2\pi \,\right\} {/eq}.

Graph of the Region D under the Given Solid

Graph of the Region D under the Given Solid

Therefore, the volume of the solid under the paraboid and over the region {eq}D {/eq} is given by the integral:

{eq}\begin{align*} Volume &= \iint \limits_D ( 1 - 3x^2 - 3y^2 ) \,dA & \text{[Evaluate as iterated integrals in polar coordinates such that } dA = r \,dr \,d\theta ] \\ &= \int_0^{2\pi} \int_0^\frac{1}{\sqrt{3}} (1 - 3r^2 ) \,r \,dr \,d\theta & \text{[Since } x^2 + y^2 = r^2 ] \\ &= \int_0^{2\pi} \int_0^\frac{1}{\sqrt{3}} (r - 3r^3) \,dr \,d\theta & \text{[Integrate with respect to } r \text{]} \\ &= \int_0^{2\pi} \left. \left( \frac{r^2}{2} - \frac{3r^4}{4} \right) \right|_0^\frac{1}{\sqrt{3}} \,d\theta & \text{[Fundamental Theorem of Calculus]} \\ &= \int_0^{2\pi} \left( \left[ \frac{1}{2} \left( \frac{1}{\sqrt{3}} \right)^2 - \frac{3}{4} \left( \frac{1}{\sqrt{3}} \right)^4 \right] - 0 \right) \,d\theta \\ &= \int_0^{2\pi} \left( \frac{1}{6} - \frac{1}{12} \right) \,d\theta \\ &= \int_0^{2\pi} \frac{1}{12} \,d\theta & \text{[Integrate with respect to } \theta \text{]} \\ &= \left. \frac{1}{12} \theta \right|_0^{2\pi} & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{12} (2\pi - 0) \\ Volume &= \boxed{ \frac{\pi}{6} } & \boxed{ \text{Volume of the Solid under the Paraboloid} } \end{align*} {/eq}


3. Volume under the paraboloid {eq}z = 1 - x^2 - y^2 {/eq} and above the xy-plane and below by the plane {eq}z = 1 - y {/eq}. Let {eq}D {/eq} be the planar region on the xy-plane that is under the given solid.

Graph of the Solid Bounded by the Paraboloid and the Plane

Graph of the Solid Bounded by the Paraboloid and the Plane

Solve for the equation of the intersection of the paraboloid and the plane:

{eq}\begin{align*} &z = 1 - x^2 - y^2 = 1 - y \\ &\Rightarrow 1 - x^2 - y^2 = 1 - y \\ &\Rightarrow x^2 + y^2 = y & \text{[Equation 1]} \\ \\ &\Rightarrow x^2 + \left( y^2 - y + \frac{1}{4} \right) = \frac{1}{4} \\ &\Rightarrow x^2 + \left( y - \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^2 & \left[ \text{Equation of a circle centered at } \left( 0 , \frac{1}{2} \right) \text{ with radius } r = \frac{1}{2} \right] \\ \\ &\Rightarrow x^2 + y^2 = y & \text{[Convert into its polar form using the identity } x^2 + y^2 = (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 ] \\ &\Rightarrow r^2 = r \sin \theta \\ &\Rightarrow r = \sin \theta & \text{[Polar equation of the circle]} \end{align*} {/eq}

Similar to Item 2, since the region {eq}D {/eq} is a circular region, it is better to define the region in polar coordinates such that {eq}D = \left\{\, (r,\theta) \,|\, 0 \le r \le \sin \theta \text{ and } 0 \le \theta \le 2\pi \,\right\} {/eq}.

Graph of the Region D on the XY-Plane under the Bounded Solid

Graph of the Region D on the XY-Plane under the Bounded Solid

Therefore, the volume of the bounded solid over the region {eq}D {/eq} is given by the integral:

{eq}\begin{align*} Volume &= \iint \limits_D \left[ ( 1 - x^2 - y^2 ) - ( 1 - y ) \right] \,dA & \text{[Convert the integrand into its polar form]} \\ &= \iint \limits_D ( y - x^2 - y^2 ) \,dA \\ &= \iint \limits_D ( r \sin \theta - r^2 ) \,dA & \text{[Evaluate as iterated integrals in polar coordinates]} \\ &= \int_0^{2\pi} \int_0^{\sin\theta} (r \sin \theta - r^2) \,r \,dr \,d\theta \\ &= \int_0^{2\pi} \int_0^{\sin\theta} (r^2 \sin \theta - r^3) \,dr \,d\theta & \text{[Integrate with respect to } r \text{]} \\ &= \int_0^{2\pi} \left. \left( \frac{r^3}{3} \sin \theta - \frac{r^4}{4} \right) \right|_0^{\sin\theta} \,d\theta & \text{[Fundamental Theorem of Calculus]} \\ &= \int_0^{2\pi} \left[ \left( \frac{\sin^3 \theta}{3} \sin \theta - \frac{\sin^4 \theta}{4} \right) - 0 \right] \,d\theta \\ &= \int_0^{2\pi} \frac{1}{12} \sin^4 \theta \,d\theta \\ &= \int_0^{2\pi} \frac{1}{12} \left( \sin^2 \theta \right)^2 \,d\theta & \left[ \text{Use the double-angle identity } \sin^2 \theta = \frac{1}{2} (1 - \cos 2\theta) \right] \\ &= \int_0^{2\pi} \frac{1}{48} \left[ \frac{1}{2} (1 - \cos 2\theta) \right]^2 \,d\theta \\ &= \int_0^{2\pi} \frac{1}{48} (1 - \cos 2\theta)^2 \,d\theta \\ &= \int_0^{2\pi} \frac{1}{48} (\cos^2 2\theta - 2 \cos 2\theta + 1) \,d\theta & \left[ \text{Use the double-angle identity } \cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta) \right] \\ &= \int_0^{2\pi} \frac{1}{48} \left( \frac{1}{2} + \frac{1}{2} \cos 4\theta - 2 \cos 2\theta + 1 \right) \,d\theta \\ &= \int_0^{2\pi} \frac{1}{96} ( \cos 4\theta - 4 \cos 2\theta + 3 ) \,d\theta & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= \frac{1}{96} \int_0^{2\pi} ( \cos 4\theta - 4 \cos 2\theta + 3 ) \,d\theta & \text{[Integrate with respect to } \theta \text{]} \\ &= \frac{1}{96} \left. \left( \frac{1}{4} \sin 4\theta - 2 \sin 2\theta + 3\theta \right) \right|_0^{2\pi} & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{96} \left( \left[ \frac{1}{4} \sin 4 (2\pi) - 2 \sin 2(2\pi) + 3(2\pi) \right] - \left[ \frac{1}{4} \sin 0 - 2 \sin 0 + 0 \right] \right) & \text{[Simplify]} \\ &= \frac{1}{96} \left[ \left( 0 - 0 + 6\pi \right) - 0 \right] \\ Volume &= \boxed{ \frac{\pi}{16} } & \boxed{ \text{Volume of the Bounded Solid} } \end{align*} {/eq}


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Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
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