Let's solve the problem **step-by-step**: --- ### **Step 1: Write Down the Surface Area Formula** When a curve \( y = f(x) \) is rotated about the \( x \)-axis, the surface area \( S \) is given by: \[ S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \] But the curve given is: \[ x^2 + y^2 = (2x^2 + 2y^2 - x)^2 \] --- ### **Step 2: Simplify the Curve Equation** Let’s try to solve for \( y \) in terms of \( x \): Let \( r^2 = x^2 + y^2 \). So, \[ r^2 = (2x^2 + 2y^2 - x)^2 \] \[ r = |2x^2 + 2y^2 - x| \] But \( y^2 = r^2 - x^2 \), so substitute back: \[ r = |2x^2 + 2(r^2 - x^2) - x| = |2x^2 + 2r^2 - 2x^2 - x| = |2r^2 - x| \] So: \[ r = |2r^2 - x| \] There are two cases: 1. \( r = 2r^2 - x \) 2. \( r = -(2r^2 - x) \Rightarrow r = -2r^2 + x \) Let's use the first case: \[ r = 2r^2 - x \implies 2r^2 - r - x = 0 \] This is a quadratic in \( r \): \[ 2r^2 - r - x = 0 \] Use the quadratic formula: \[ r = \frac{1 \pm \sqrt{1 + 8x}}{4} \] Recall \( r^2 = x^2 + y^2 \), so: \[ x^2 + y^2 = r^2 = \left(\frac{1 \pm \sqrt{1 + 8x}}{4}\right)^2 \] Thus, \[ y^2 = \left(\frac{1 \pm \sqrt{1 + 8x}}{4}\right)^2 - x^2 \] --- ### **Step 3: Find the Interval of \( x \)** For \( y^2 \geq 0 \), need: \[ \left(\frac{1 \pm \sqrt{1 + 8x}}{4}\right)^2 \geq x^2 \] Let’s use \( r = \frac{1 + \sqrt{1 + 8x}}{4} \) (since \( r \geq 0 \)), so: \[ \frac{1 + \sqrt{1 + 8x}}{4} \geq 0 \implies 1 + \sqrt{1 + 8x} \geq 0 \implies \sqrt{1 + 8x} \geq -1 \] But since \( \sqrt{1 + 8x} \geq 0 \), so \( x \geq -\frac{1}{8} \). Also, for \( y^2 \geq 0 \): \[ \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 \geq x^2 \] Let’s solve this inequality: Let \( s = \sqrt{1 + 8x} \): \[ \left(\frac{1 + s}{4}\right)^2 \geq x^2 \] \[ \frac{(1 + s)^2}{16} \geq x^2 \] \[ (1 + s)^2 \geq 16 x^2 \] \[ 1 + 2s + s^2 \geq 16 x^2 \] But \( s^2 = 1 + 8x \), so: \[ 1 + 2s + (1 + 8x) \geq 16 x^2 \] \[ 2 + 2s + 8x \geq 16 x^2 \] \[ 16 x^2 - 8x - 2 - 2s \leq 0 \] This is a bit messy, but for the sake of the problem, let's consider the range where \( y \) is real (i.e., the curve exists). --- ### **Step 4: Find \( \frac{dy}{dx} \)** Let’s focus on the positive root: \[ y^2 = r^2 - x^2 = \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 - x^2 \] So \[ y = \sqrt{ \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 - x^2 } \] Let’s denote \( u = \frac{1 + \sqrt{1 + 8x}}{4} \), so \( y = \sqrt{u^2 - x^2} \). Compute \( \frac{dy}{dx} \): Let’s write: \[ y = \sqrt{u^2 - x^2} \] where \( u = \frac{1 + \sqrt{1 + 8x}}{4} \). Then, \[ \frac{dy}{dx} = \frac{1}{2\sqrt{u^2 - x^2}} \cdot \left[2u \frac{du}{dx} - 2x \right] \] \[ = \frac{u \frac{du}{dx} - x}{\sqrt{u^2 - x^2}} \] Now, \[ \frac{du}{dx} = \frac{1}{4} \cdot \frac{1}{2\sqrt{1 + 8x}} \cdot 8 = \frac{1}{4} \cdot \frac{8}{2\sqrt{1 + 8x}} = \frac{2}{\sqrt{1 + 8x}} \] So, \[ \frac{dy}{dx} = \frac{u \cdot \frac{2}{\sqrt{1 + 8x}} - x}{\sqrt{u^2 - x^2}} \] --- ### **Step 5: Write the Surface Area Integral** \[ S = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \] We already have \( y \) and \( \frac{dy}{dx} \). The limits \( x_1 \) and \( x_2 \) should be where \( y = 0 \), i.e., where: \[ \left( \frac{1 + \sqrt{1 + 8x}}{4} \right)^2 - x^2 = 0 \] Set \( y = 0 \): \[ \left( \frac{1 + \sqrt{1 + 8x}}{4} \right)^2 = x^2 \] \[ \frac{1 + \sqrt{1 + 8x}}{4} = \pm x \] First, for \( x > 0 \): \[ \frac{1 + \sqrt{1 + 8x}}{4} = x \] \[ 1 + \sqrt{1 + 8x} = 4x \] \[ \sqrt{1 + 8x} = 4x - 1 \] \[ 1 + 8x = (4x - 1)^2 \] \[ 1 + 8x = 16x^2 - 8x + 1 \] \[ 0 = 16x^2 - 16x \] \[ 16x(x - 1) = 0 \implies x = 0 \text{ or } x = 1 \] For \( x < 0 \): \[ \frac{1 + \sqrt{1 + 8x}}{4} = -x \] \[ 1 + \sqrt{1 + 8x} = -4x \] \[ \sqrt{1 + 8x} = -4x - 1 \] But \( \sqrt{1 + 8x} \geq 0 \), so \( -4x - 1 \geq 0 \implies x \leq -\frac{1}{4} \) Square both sides: \[ 1 + 8x = ( -4x - 1 )^2 = 16x^2 + 8x + 1 \] \[ 1 + 8x = 16x^2 + 8x + 1 \] \[ 0 = 16x^2 \implies x = 0 \] So, the only valid \( x \) interval is \( x \in [0, 1] \). --- ### **Step 6: Final Surface Area Integral** \[ S = 2\pi \int_{0}^{1} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \] where \[ y = \sqrt{ \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 - x^2 } \] and \[ \frac{dy}{dx} = \frac{u \cdot \frac{2}{\sqrt{1 + 8x}} - x}{\sqrt{u^2 - x^2}}, \quad u = \frac{1 + \sqrt{1 + 8x}}{4} \] --- ### **Step 7: Final Answer** \[ \boxed{ S = 2\pi \int_{0}^{1} \sqrt{ \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 - x^2 } \, \sqrt{1 + \left[ \frac{ \frac{1 + \sqrt{1 + 8x}}{4} \cdot \frac{2}{\sqrt{1 + 8x}} - x }{ \sqrt{ \left(\frac{1 + \sqrt{1 + 8x}}{4}\right)^2 - x^2 } } \right]^2 } \, dx } \] This is the explicit **surface area** in terms of a definite integral over \( x \) from \( 0 \) to \( 1 \). --- #### If you need a numerical value, this integral can be evaluated with a calculator or using computational tools. Let me know if you'd like an approximate value!

Certainly! Let's carefully go through the problem step-by-step, providing detailed calculations and reasoning at each stage. --- ## **Problem Restatement** Find the surface area of the surface generated when the curve defined by the given relation is revolved about the z-axis. The given relation (from the image) appears to be: \[ x^2 + y^2 = (2x^2 + 2y^2 - x)^2 \] --- ## **Step 1: Simplify the curve equation** Let’s rewrite the equation: \[ x^2 + y^2 = (2x^2 + 2 y^2 - x)^2 \] Define: \[ r^2 = x^2 + y^2 \] Substituting: \[ r^2 = (2x^2 + 2 y^2 - x)^2 \] Express the right side: \[ r = \pm (2x^2 + 2 y^2 - x) \] But since \( r \geq 0 \), we take: \[ r = |2x^2 + 2 y^2 - x| \] --- ## **Step 2: Express \( r \) in terms of \( x \) and \( y \)** Recall: \[ r^2 = x^2 + y^2 \] and \[ r = |2x^2 + 2 y^2 - x| \] which leads to two cases: ### **Case 1:** \[ r = 2x^2 + 2 y^2 - x \] Substitute \( r^2 = x^2 + y^2 \): \[ r = 2x^2 + 2 r^2 - x \] Rearranged: \[ r - 2 r^2 = 2 x^2 - x \] or: \[ -2 r^2 + r = 2 x^2 - x \] Multiply both sides by \(-1\): \[ 2 r^2 - r = -2 x^2 + x \] --- ### **Case 2:** \[ r = -(2x^2 + 2 y^2 - x) = -2x^2 - 2 y^2 + x \] Similarly, substitute \( r^2 = x^2 + y^2 \): \[ r = -2x^2 - 2 r^2 + x \] Rearranged: \[ r + 2 r^2 = -2 x^2 + x \] Multiply both sides by 1 (no change): \[ r + 2 r^2 = -2 x^2 + x \] --- ## **Step 3: Solve for \( r \) in each case** Let's analyze **Case 1**: \[ 2 r^2 - r = -2 x^2 + x \] This is a quadratic in \( r \): \[ 2 r^2 - r + (2 x^2 - x) = 0 \] Quadratic form: \[ 2 r^2 - r + (2 x^2 - x) = 0 \] Solve for \( r \): \[ r = \frac{1 \pm \sqrt{1 - 8 (2 x^2 - x)}}{4} \] Because: \[ \Delta = 1 - 8 (2 x^2 - x) = 1 - 16 x^2 + 8 x \] Similarly, for **Case 2**: \[ 2 r^2 + r + (2 x^2 - x) = 0 \] Quadratic: \[ 2 r^2 + r + (2 x^2 - x) = 0 \] Solution: \[ r = \frac{-1 \pm \sqrt{1 - 8 (2 x^2 - x)}}{4} \] Note that in both cases, the discriminant is: \[ \Delta = 1 - 16 x^2 + 8 x \] --- ## **Step 4: Find the domain of \( x \)** For real solutions: \[ \Delta \geq 0 \] \[ 1 - 16 x^2 + 8 x \geq 0 \] Rewrite: \[ -16 x^2 + 8 x + 1 \geq 0 \] Divide through by \(-1\): \[ 16 x^2 - 8 x - 1 \leq 0 \] Quadratic in \( x \): \[ 16 x^2 - 8 x - 1 \leq 0 \] Find roots: \[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 16 \times (-1)}}{2 \times 16} = \frac{8 \pm \sqrt{64 + 64}}{32} = \frac{8 \pm \sqrt{128}}{32} \] \[ \sqrt{128} = 8 \sqrt{2} \] So: \[ x = \frac{8 \pm 8 \sqrt{2}}{32} = \frac{1 \pm \sqrt{2}}{4} \] Since the quadratic opens upward (coefficient of \( x^2 \) is positive), the inequality holds between the roots: \[ x \in \left[ \frac{1 - \sqrt{2}}{4}, \frac{1 + \sqrt{2}}{4} \right] \] Numerically: \[ \frac{1 - 1.4142}{4} \approx \frac{-0.4142}{4} \approx -0.10355 \] \[ \frac{1 + 1.4142}{4} \approx \frac{2.4142}{4} \approx 0.60355 \] --- ## **Step 5: Select the relevant branch** Because \( r = \sqrt{x^2 + y^2} \geq 0 \), and from earlier considerations, the physically meaningful solutions are obtained when: \[ r = \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \] which corresponds to the positive root in **Case 1**. --- ## **Step 6: Express \( y \) in terms of \( x \)** Recall: \[ r^2 = x^2 + y^2 \] and \[ r = \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \] Thus: \[ x^2 + y^2 = r^2 = \left( \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \right)^2 \] So, \[ y = \pm \sqrt{ r^2 - x^2 } = \pm \sqrt{ \left( \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \right)^2 - x^2 } \] Since the surface area involves the magnitude of \( y \), we can take the positive branch: \[ y = \sqrt{ \left( \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \right)^2 - x^2 } \] --- ## **Step 7: Calculate \( dy/dx \)** Let: \[ u(x) = \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \] then: \[ y = \sqrt{ u^2 - x^2 } \] Differentiate: \[ \frac{dy}{dx} = \frac{1}{2 \sqrt{u^2 - x^2}} \left( 2 u \frac{du}{dx} - 2 x \right ) \] simplify: \[ \frac{dy}{dx} = \frac{ u \frac{du}{dx} - x }{ \sqrt{ u^2 - x^2 } } \] Now, find \( du/dx \): \[ u = \frac{1 + \sqrt{1 - 8(2x^2 - x)}}{4} \] Let: \[ w(x) = 1 - 8(2x^2 - x) = 1 - 16 x^2 + 8 x \] Then: \[ u = \frac{1 + \sqrt{w(x)}}{4} \] Calculate: \[ \frac{du}{dx} = \frac{1}{4} \times \frac{1}{2 \sqrt{w(x)}} \times w'(x) \] where: \[ w'(x) = -32 x + 8 \] Thus: \[ \frac{du}{dx} = \frac{1}{8 \sqrt{w(x)}} \times (-32 x + 8) = \frac{-32 x + 8}{8 \sqrt{w(x)}} = \frac{-4 x + 1}{ \sqrt{w(x)} } \] Note: \[ w(x) = 1 - 16 x^2 + 8 x \] --- ## **Step 8: Write \( dy/dx \) explicitly** Now: \[ \frac{dy}{dx} = \frac{ u \times \frac{-4 x + 1}{ \sqrt{w(x)} } - x }{ \sqrt{ u^2 - x^2 } } \] with \[ u = \frac{1 + \sqrt{w(x)}}{4} \] and \[ w(x) = 1 - 16 x^2 + 8 x \] --- ## **Step 9: Surface area integral** The surface area generated by revolving the curve around the z-axis is: \[ S = 2 \pi \int_{x = a}^{b} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx \] where: - \( a = \frac{1 - \sqrt{2}}{4} \approx -0.10355 \), - \( b = \frac{1 + \sqrt{2}}{4} \approx 0.60355 \), - \( y = \sqrt{ u^2 - x^2 } \), - \( \frac{dy}{dx} \) as above. --- ## **Final Answer:** \[ \boxed{ \text{Surface area } S = 2 \pi \int_{x = \frac{1 - \sqrt{2}}{4}}^{\frac{1 + \sqrt{2}}{4}} y \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } dx } \] where: \[ \begin{aligned} & y = \sqrt{ u^2 - x^2 } \quad \text{with} \quad u = \frac{1 + \sqrt{w(x)}}{4} \\ & w(x) = 1 - 16 x^2 + 8 x \\ & \frac{dy}{dx} = \frac{ u \times \frac{-4 x + 1}{ \sqrt{w(x)} } - x }{ \sqrt{ u^2 - x^2 } } \end{aligned} \] --- ## **Remarks:** - The integral can be evaluated numerically using software like WolframAlpha, MATLAB, or Python. - The key steps involved simplifying the original relation, identifying the domain, expressing \( y \) and \( dy/dx \), and setting up the integral for the surface area. --- If you want a numerical approximation or further simplification, please let me know!