Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

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Asked on July 28, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-07-28T18:38:44.646Z

6.1 Calculate the length of the curve given by x = e^2 and y = e^2 for 0 ≤ ≤ . Assumption: Based on typical problem complexity for this type of question, it is assumed that the second function is y = e^2 , as y = e^2 would lead to a significantly more complex integral. Step 1: Find the derivatives (dx)/(d) and (dy)/(d). Given x = e^2 : (dx)/(d) = (d)/(d)(e^2 ) = (2e^2) + e^2(-) = e^2(2 - ) Given y = e^2 : (dy)/(d) = (d)/(d)(e^2 ) = (2e^2) + e^2() = e^2(2 + ) Step 2: Calculate ((dx)/(d))^2 + ((dy)/(d))^2. ((dx)/(d))^2 = (e^2(2 - ))^2 = e^4(4^2 - 4 + ^2) ((dy)/(d))^2 = (e^2(2 + ))^2 = e^4(4^2 + 4 + ^2) Summing these: ((dx)/(d))^2 + ((dy)/(d))^2 = e^4(4^2 - 4 + ^2 + 4^2 + 4 + ^2) = e^4(5^2 + 5^2) = 5e^4(^2 + ^2) = 5e^4(1) = 5e^4 Step 3: Set up the arc length integral. The arc length formula for parametric curves is L = _^ sqrt(((dx)/(d))^2 + ((dy)/(d))^2) d. L = _0^ sqrt(5e^4) d = _0^ sqrt(5) sqrt(e^4) d = _0^ sqrt(5) e^2 d Step 4: Evaluate the integral. L = sqrt(5) [ (1)/(2)e^2 ]_0^ L = sqrt(5)2 (e^2 - e^2(0)) L = sqrt(5)2 (e^2 - e^0) L = sqrt(5)2 (e^2 - 1) The length of the curve is sqrt(5)2 (e^2 - 1). 6.2 Calculate the surface area generated when the curve of y = x between x = 0 and x = ()/(2) rotates about the x-axis. Step 1: Find the derivative (dy)/(dx). Given y = x: (dy)/(dx) = (d)/(dx)( x) = - x Step 2: Calculate sqrt(1 + ((dy)/(dx))^2). sqrt(1 + (- x)^2) = sqrt(1 + ^2 x) Step 3: Set up the surface area integral. The surface area of revolution about the x-axis is S = _a^b 2 y sqrt(1 + ((dy)/(dx))^2) dx. S = _0^()/(2) 2 ( x) sqrt(1 + ^2 x) dx Step 4: Use substitution to evaluate the integral. Let u = x. Then du = x \, dx. When x = 0, u = (0) = 0. When x = ()/(2), u = (()/(2)) = 1. The integral becomes: S = 2 _0^1 sqrt(1 + u^2) du This is a standard integral of the form sqrt(a^2 + u^2) du = (u)/(2)sqrt(a^2+u^2) + (a^2)/(2)|u+sqrt(a^2+u^2)|. Here a=1. S = 2 [ (u)/(2)sqrt(1+u^2) + (1)/(2)|u+sqrt(1+u^2)| ]_0^1 Step 5: Evaluate the definite integral. Evaluate at the upper limit u=1: (1)/(2)sqrt(1+1^2) + (1)/(2)|1+sqrt(1+1^2)| = (1)/(2)sqrt(2) + (1)/(2)(1+sqrt(2)) Evaluate at the lower limit u=0: (0)/(2)sqrt(1+0^2) + (1)/(2)|0+sqrt(1+0^2)| = 0 + (1)/(2)(1) = 0 Subtract the lower limit value from the upper limit value: S = 2 ( (sqrt(2)2 + (1)/(2)(1+sqrt(2))) - 0 ) S = 2 ( sqrt(2)2 + (1)/(2)(1+sqrt(2)) ) S = (2) + (1+sqrt(2)) The surface area generated is (2) + (1+sqrt(2)). Last free one today — make it count tomorrow, or type /upgrade for unlimited.

Accepted Answer · 2026-07-28T18:38:44.646Z

6.1 Calculate the length of the curve given by x = e^2 and y = e^2 for 0 ≤ ≤ . Assumption: Based on typical problem complexity for this type of question, it is assumed that the second function is y = e^2 , as y = e^2 would lead to a significantly more complex integral. Step 1: Find the derivatives (dx)/(d) and (dy)/(d). Given x = e^2 : (dx)/(d) = (d)/(d)(e^2 ) = (2e^2) + e^2(-) = e^2(2 - ) Given y = e^2 : (dy)/(d) = (d)/(d)(e^2 ) = (2e^2) + e^2() = e^2(2 + ) Step 2: Calculate ((dx)/(d))^2 + ((dy)/(d))^2. ((dx)/(d))^2 = (e^2(2 - ))^2 = e^4(4^2 - 4 + ^2) ((dy)/(d))^2 = (e^2(2 + ))^2 = e^4(4^2 + 4 + ^2) Summing these: ((dx)/(d))^2 + ((dy)/(d))^2 = e^4(4^2 - 4 + ^2 + 4^2 + 4 + ^2) = e^4(5^2 + 5^2) = 5e^4(^2 + ^2) = 5e^4(1) = 5e^4 Step 3: Set up the arc length integral. The arc length formula for parametric curves is L = _^ sqrt(((dx)/(d))^2 + ((dy)/(d))^2) d. L = _0^ sqrt(5e^4) d = _0^ sqrt(5) sqrt(e^4) d = _0^ sqrt(5) e^2 d Step 4: Evaluate the integral. L = sqrt(5) [ (1)/(2)e^2 ]_0^ L = sqrt(5)2 (e^2 - e^2(0)) L = sqrt(5)2 (e^2 - e^0) L = sqrt(5)2 (e^2 - 1) The length of the curve is sqrt(5)2 (e^2 - 1). 6.2 Calculate the surface area generated when the curve of y = x between x = 0 and x = ()/(2) rotates about the x-axis. Step 1: Find the derivative (dy)/(dx). Given y = x: (dy)/(dx) = (d)/(dx)( x) = - x Step 2: Calculate sqrt(1 + ((dy)/(dx))^2). sqrt(1 + (- x)^2) = sqrt(1 + ^2 x) Step 3: Set up the surface area integral. The surface area of revolution about the x-axis is S = _a^b 2 y sqrt(1 + ((dy)/(dx))^2) dx. S = _0^()/(2) 2 ( x) sqrt(1 + ^2 x) dx Step 4: Use substitution to evaluate the integral. Let u = x. Then du = x \, dx. When x = 0, u = (0) = 0. When x = ()/(2), u = (()/(2)) = 1. The integral becomes: S = 2 _0^1 sqrt(1 + u^2) du This is a standard integral of the form sqrt(a^2 + u^2) du = (u)/(2)sqrt(a^2+u^2) + (a^2)/(2)|u+sqrt(a^2+u^2)|. Here a=1. S = 2 [ (u)/(2)sqrt(1+u^2) + (1)/(2)|u+sqrt(1+u^2)| ]_0^1 Step 5: Evaluate the definite integral. Evaluate at the upper limit u=1: (1)/(2)sqrt(1+1^2) + (1)/(2)|1+sqrt(1+1^2)| = (1)/(2)sqrt(2) + (1)/(2)(1+sqrt(2)) Evaluate at the lower limit u=0: (0)/(2)sqrt(1+0^2) + (1)/(2)|0+sqrt(1+0^2)| = 0 + (1)/(2)(1) = 0 Subtract the lower limit value from the upper limit value: S = 2 ( (sqrt(2)2 + (1)/(2)(1+sqrt(2))) - 0 ) S = 2 ( sqrt(2)2 + (1)/(2)(1+sqrt(2)) ) S = (2) + (1+sqrt(2)) The surface area generated is (2) + (1+sqrt(2)). Last free one today — make it count tomorrow, or type /upgrade for unlimited.

Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

6.1 Calculate the length of the curve given by $x = e^{2\theta} \cos\theta$ and $y = e^{2\theta} \sin\theta$ for $0 \leq \theta \leq \pi$ .

6.2 Calculate the surface area generated when the curve of $y = \cos x$ between $x = 0$ and $x = \frac{\pi}{2}$ rotates about the x-axis.

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Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

6.1 Calculate the length of the curve given by $x = e^{2\theta} \cos\theta$ and $y = e^{2\theta} \sin\theta$ for $0 \leq \theta \leq \pi$ .

6.2 Calculate the surface area generated when the curve of $y = \cos x$ between $x = 0$ and $x = \frac{\pi}{2}$ rotates about the x-axis.

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Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

6.1 Calculate the length of the curve given by x=e2θcos⁡θx = e^{2\theta} \cos\thetax=e2θcosθ and y=e2θsin⁡θy = e^{2\theta} \sin\thetay=e2θsinθ for 0≤θ≤π0 \leq \theta \leq \pi0≤θ≤π.

6.2 Calculate the surface area generated when the curve of y=cos⁡xy = \cos xy=cosx between x=0x = 0x=0 and x=π2x = \frac{\pi}{2}x=2π​ rotates about the x-axis.

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Calculate the length of the curve given by x = e2theta costheta and y = e2theta sintheta for 0 theta pi. Calculate the surface area generated when the curve of y = cosx between x = 0 and x = (pi)/(2) rotates about the x-axis.

6.1 Calculate the length of the curve given by x=e2θcos⁡θx = e^{2\theta} \cos\thetax=e2θcosθ and y=e2θsin⁡θy = e^{2\theta} \sin\thetay=e2θsinθ for 0≤θ≤π0 \leq \theta \leq \pi0≤θ≤π.

6.2 Calculate the surface area generated when the curve of y=cos⁡xy = \cos xy=cosx between x=0x = 0x=0 and x=π2x = \frac{\pi}{2}x=2π​ rotates about the x-axis.

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6.1 Calculate the length of the curve given by $x = e^{2\theta} \cos\theta$ and $y = e^{2\theta} \sin\theta$ for $0 \leq \theta \leq \pi$ .

6.2 Calculate the surface area generated when the curve of $y = \cos x$ between $x = 0$ and $x = \frac{\pi}{2}$ rotates about the x-axis.

6.1 Calculate the length of the curve given by $x = e^{2\theta} \cos\theta$ and $y = e^{2\theta} \sin\theta$ for $0 \leq \theta \leq \pi$ .

6.2 Calculate the surface area generated when the curve of $y = \cos x$ between $x = 0$ and $x = \frac{\pi}{2}$ rotates about the x-axis.