Copyright

Find the volume of the solid that bounded by i) z = x^2 + y^2 and z = 1 - x^2. ii) 4 x^2 + y^2 =...

Question:

Find the volume of the solid that bounded by

i) {eq}z = x^2 + y^2 {/eq} and {eq}z = 1 - x^2 {/eq}.

ii) {eq}4 x^2 + y^2 = 4 {/eq} and {eq}z + y = 2 {/eq}.

Volumes of Solids:

The volume of the solid {eq}S {/eq} in {eq}\mathbb{R}^3 {/eq} can be determined by evaluating the triple integrals:

$$Volume = \iiint \limits_S dV $$

Suppose the solid {eq}S {/eq} is bounded by the smooth surfaces of the functions {eq}F(x,y) \le z \le G(x,y) {/eq} over the planar region {eq}D {/eq} which lies on the xy-plane.

Then, the triple integrals for the volume of the solid can be written as the double integrals:

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

Furthermore, suppose the planar region {eq}D {/eq} can be defined as a Type I region, one which is bounded by continuous functions of x, such that the region is described as {eq}h_1(x) \le y h_2(x) {/eq} where {eq}a \le x \le b {/eq}.

The double integrals for the volume of the solid can then be expressed as the iterated integrals:

$$Volume = \int_a^b \int_{h_1(x)}^{h_2(x)} \left[ G(x,y) - F(x,y) \right] dy \,dx $$

The same principle can be applied to when the region {eq}D {/eq} is a Type II region, a planar region which is instead bounded by continuous functions of y over some interval/s of y.

Answer and Explanation:

1. Let {eq}S1 {/eq} be the solid bounded by {eq}z = x^2 + y^2 {/eq} and {eq}z = 1 - x^2 {/eq}.

Graph of the Solid S1

Graph of the Solid S1

Additionally, let {eq}D1 {/eq} be the region on the xy-plane that is directly under the solid {eq}S1 {/eq}. Solve for the equation of the intersection of the two surfaces:

{eq}\begin{align*} &z = x^2 + y^2 = 1 - x^2 \\ &\Rightarrow x^2 + y^2 = 1 - x^2 \\ &\Rightarrow 2x^2 + y^2 = 1 & \text{[Equation of the Elliptical Region } D1 ] \end{align*} {/eq}

Graph of the Region D1 under the Solid S1

Graph of the Region D1 under the Solid S1

Based on the graphs, {eq}z = 1 - x^2 \ge z = x^2 + y^2 {/eq}. This can be confirmed algebraically by taking a point in the region {eq}D {/eq} and solving for the value {eq}z {/eq} using the equations of the two surfaces.

For instance, the origin {eq}(0,0) {/eq} is in the region {eq}D {/eq}, and {eq}1 - (0)^2 = 1 > 0 = (0)^2 + (0)^2 {/eq}.


Moving on to the region of integration, express the equation of the ellipse as functions of x to define the region {eq}D1 {/eq} as a Type I region.

{eq}\begin{align*} &2x^2 + y^2 = 1 \\ &\Rightarrow y^2 = 1 - 2x^2 \\ &\Rightarrow y = \sqrt{ 1 - 2x^2 } & \text{[Equation of the Upper Semi-Ellipse]} \\ &\Rightarrow y = - \sqrt{ 1 - 2x^2 } & \text{[Equation of the Lower Semi-Ellipse]} \\ \\ &y^2 = 1 - 2x^2 = 0 & \text{[Solve for the } x \text{-intercepts]} \\ &\Rightarrow 2x^2 = 1 \\ &\Rightarrow x^2 = \frac{1}{2} \\ &\Rightarrow x = \pm \frac{1}{\sqrt{2}} \end{align*} {/eq}

Therefore, the solid {eq}S1 {/eq} can be defined as {eq}x^2 + y^2 \le z \le 1 - x^2 {/eq} over the region {eq}D1 = \left\{\, (x,y) \,|\, -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \text{ and } -\sqrt{ 1 - 2x^2 } \le y \le \sqrt{ 1 - 2x^2 } \,\right\} {/eq}.

Finally, the volume of the solid {eq}S1 {/eq} is given by the double integrals:

{eq}\begin{align*} Volume_{S1} &= \iint \limits_{D1} [ (1 - x^2) - (x^2 + y^2) ] \,dA & \text{[Evaluate as iterated integrals]} \\ &= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \int_{-\sqrt{ 1 - 2x^2 }}^{\sqrt{ 1 - 2x^2 }} ( 1 - 2x^2 - y^2 ) \,dy \,dx & \text{[Integrate with respect to } y \text{]} \\ &= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \left. \left( y - 2x^2 y - \frac{y^3}{3} \right) \right|_{-\sqrt{ 1 - 2x^2 }}^{\sqrt{ 1 - 2x^2 }} \,dx & \text{[Note that the result of the integral is an odd function. Thus, } f(-a) = - f(a) \text{ and } f(a) - f(-a) = 2f(a) ] \\ &= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 2 \left( \sqrt{ 1 - 2x^2 } - 2x^2 \sqrt{ 1 - 2x^2 } - \frac{\sqrt{ 1 - 2x^2 }^3}{3} \right) \,dx \\ &= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 2 \sqrt{ 1 - 2x^2 } \left( 1 - 2x^2 - \frac{1 - 2x^2}{3} \right) \,dx \\ &= \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \frac{4}{3} (1 - 2x^2)^\frac{3}{2} \,dx & \left[ \text{Substitute } x = \frac{1}{\sqrt{2}} \cos w \text{ and } dx = \frac{-1}{\sqrt{2}} \sin w \,dw \right] \\ &= \int_{w\left(-\frac{1}{\sqrt{2}}\right)}^{w\left(\frac{1}{\sqrt{2}}\right)} \frac{4}{3} \left[ 1 - 2 \left( \frac{1}{\sqrt{2}} \cos w \right)^2 \right]^\frac{3}{2} \left( \frac{-1}{\sqrt{2}} \sin w \,dw \right) & \left[ \text{The new limits of integration are } w\left(\frac{-1}{\sqrt{2}}\right) = \cos^{-1} \sqrt{2} \left(\frac{-1}{\sqrt{2}}\right) = \cos^{-1} (-1) = \pi \text{ and } w\left(-\frac{1}{\sqrt{2}}\right) = \cos^{-1} \sqrt{2} \left(\frac{1}{\sqrt{2}}\right) = \cos^{-1} 1 = 0 \right] \\ &= \int_\pi^0 - \frac{4}{3\sqrt{2}} \sin w ( 1 - \cos^2 w )^\frac{3}{2} \,dw & \text{[Use the identity } \cos^2 \theta + \sin^2 \theta = 1 ] \\ &= \int_\pi^0 - \frac{4}{3\sqrt{2}} \sin w ( \sin^2 w )^\frac{3}{2} \,dw & \left[ \int_a^b f(x) \,dx = \int_b^a - f(x) \,dx \right] \\ &= \int_0^\pi \frac{4}{3\sqrt{2}} \sin w \sin^3 w \,dw & \left[ \text{For all constant } k \,,\, \int k \cdot f(x) \,dx = k \int f(x)\,dx \right] \\ &= \frac{4}{3\sqrt{2}} \int_0^\pi \sin^3 w \sin w \,dw & \text{[Integrate by parts. Let } u = \sin^3 w \text{ and } dv = \sin w \,dw \text{ such that } du = 3 \sin^2 w \cos w \,dw \text{ and } v = - \cos w ] \\ &= \frac{4}{3\sqrt{2}} \left[ \left. \left( - \cos w \sin^3 w \right) \right|_0^\pi - \int_0^\pi - 3 \sin^2 w \cos^2 w \,dw \right] & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{4}{3\sqrt{2}} \left[ ( 0 - 0 ) + 3 \int_0^\pi \sin^2 w \cos^2 w \,dw \right] & \left[ \text{Use the identity } \sin^2 \theta \cos^2 \theta = \frac{1}{8} (1 - \cos 4\theta) \right] \\ &= \frac{4}{\sqrt{2}} \int_0^\pi \frac{1}{8} (1 - \cos 4w) \,dw & \text{[Substitute } u = 4w \text{ such that } du = 4\,dw \,,\, u(0) = 0 \text{ and } u(\pi) = 4\pi ] \\ &= \frac{4}{\sqrt{2}} \int_0^{4\pi} \frac{1}{32} (1 - \cos u) \,du \\ &= \frac{1}{8\sqrt{2}} \int_0^{4\pi} (1 - \cos u) \,du & \text{[Integrate with respect to } u \text{]} \\ &= \frac{1}{8\sqrt{2}} \left. \left( u - \sin u \right) \right|_0^{4\pi} & \text{[Fundamental Theorem of Calculus]} \\ &= \frac{1}{8\sqrt{2}} \left[ ( 4\pi - \sin 4\pi ) - ( 0 - \sin 0 ) \right] \\ &= \frac{1}{8\sqrt{2}} ( 4\pi ) \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ Volume_{S1} &= \boxed{ \frac{\pi \sqrt{2}}{4} } & \boxed{ \text{Volume of the Solid } S1 } \end{align*} {/eq}


2. Let {eq}S2 {/eq} be the solid bounded by {eq}4x^2 + y^2 = 4 {/eq} and {eq}z + y = 2 {/eq}. We will assume that the problem pertains to the portion of the solid above the xy-plane {eq}(z=0) {/eq} that is bounded by the graphs of the given equations.

Graph of the Solid S2

Graph of the Solid S2

Based on the graph, solid is bounded above by the plane {eq}z = 2 - y {/eq} and below by the region {eq}D {/eq} on the xy-plane {eq}(z=0) {/eq}, which is in turn bounded by the ellipse {eq}4x^2 + y^2 = 4 {/eq}.

Again, defining the region of integration as a Type I region, express the equation of the ellipse as functions of x:

{eq}\begin{align*} &4x^2 + y^2 = 4 \\ &\Rightarrow y^2 = 4 - 4x^2 \\ &\Rightarrow y = 2 \sqrt{ 1 - x^2 } & \text{[Equation of the Upper Semi-Ellipse]} \\ &\Rightarrow y = - 2 \sqrt{ 1 - x^2 } & \text{[Equation of the Lower Semi-Ellipse]} \\ \\ &y^2 = 4 - 4x^2 = 0 & \text{[Solve for the } x \text{-intercepts]} \\ &\Rightarrow 4x^2 = 4 \\ &\Rightarrow x^2 = 1 \\ &\Rightarrow x = \pm 1 \end{align*} {/eq}

Therefore, the solid {eq}S2 {/eq} is defined as {eq}0 \le z \le 2 - y {/eq} and over the region {eq}D2 = \left\{\, (x,y) \,|\, -1 \le x \le 1 \text{ and } -2 \sqrt{ 1 - x^2 } \le y \le 2 \sqrt{ 1 - x^2 } \,\right\} {/eq}.

Then, the volume of the solid {eq}S2 {/eq} is given by the double integrals:

{eq}\begin{align*} Volume_{S2} &= \iint \limits_{D2} [ (2 - y) - 0 ] \,dA & \text{[Evaluate as iterated integrals]} \\ &= \int_{-1}^1 \int_{-2 \sqrt{ 1 - x^2 }}^{2 \sqrt{ 1 - x^2 }} (2 - y) \,dy \,dx & \text{[Integrate with respect to } y \text{]} \\ &= \int_{-1}^1 \left. \left( 2y - \frac{y^2}{2} \right) \right|_{-2 \sqrt{ 1 - x^2 }}^{2 \sqrt{ 1 - x^2 }} & \text{[Fundamental Theorem of Calculus]} \\ &= \int_{-1}^1 \left( \left[ 2 \left( 2 \sqrt{ 1 - x^2 } \right) - \frac{1}{2} \left( 2 \sqrt{ 1 - x^2 } \right)^2 \right] - \left[ 2 \left( -2 \sqrt{ 1 - x^2 } \right) - \frac{1}{2} \left( -2 \sqrt{ 1 - x^2 } \right)^2 \right] \right) \,dx \\ &= \int_{-1}^1 \left[ \left( 4 \sqrt{1 - x^2} - 2 + 2x^2 \right) - \left( - 4 \sqrt{1 - x^2} - 2 + 2x^2 \right) \right] \,dx \\ &= \int_{-1}^1 8 \sqrt{1 - x^2} \,dx & \text{[Substitute } x = \cos u \text{ such that } dx = - \sin u \,du \,,\, u(-1) = \cos^{-1} (-1) = \pi \text{ and } u(1) = \cos^{-1} 1 = 0 ] \\ &= \int_\pi^0 -8 \sin u \sqrt{1 - \cos^2 u} \,du & \text{[Use the identity } \cos^2 \theta + \sin^2 \theta = 1 ] \\ &= \int_\pi^0 -8 \sin u \sqrt{ \sin^2 u } \,du & \left[ \int_a^b f(x) \,dx = \int_b^a - f(x) \,dx \right] \\ &= \int_0^\pi 8 \sin^2 u \,du & \left[ \text{Use the identity } \sin^2 \theta = \frac{1}{2} (1 - \cos 2\theta) \right] \\ &= \int_0^\pi 4 (1 - \cos 2u) \,du & \text{[Substitute } w = 2u \text{ such that } dw = 2\,du \,,\, w(0) = 0 \text{ and } w(\pi) = 2\pi ] \\ &= \int_0^{2\pi} 2 (1 - \cos w) \,dw \\ &= 2 \int_0^{2\pi} (1 - \cos w) \,dw & \text{[Integrate with respect to } w \text{]} \\ &= 2 \left. \left( w - \sin w \right) \right|_0^{2\pi} & \text{[Fundamental Theorem of Calculus]} \\ &= 2 [ (2\pi - \sin 2\pi) - (0 - \sin 0) ] \\ &= 2 (2\pi) \\ Volume_{S2} &= \boxed{ 4\pi } & \boxed{ \text{Volume of the Solid } S2 } \end{align*} {/eq}


Learn more about this topic:

Loading...
Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4
512

Related to this Question

Explore our homework questions and answers library