You are given that C=∫0πe−2xcos⁡x  dxC = \int_{0}^{\pi} \mathrm{e}^{-2x} \cos x \; \mathrm{d}x and S=∫0πe−2xsin⁡x  dxS = \int_{0}^{\pi} \mathrm{e}^{-2x} \sin x \; \mathrm{d}x. Use integration by parts to show C=2SandS=−2C+1+e−2πC = 2S \quad \text{and} \quad S = -2C + 1 + \mathrm{e}^{-2\pi} Hence, find the exact values of CC and SS. The finite region bounded by the axes and the curve y=e−xcos⁡(12x)y = \mathrm{e}^{-x} \cos\left(\frac{1}{2}x\right), for 0≤x≤π0 \le x \le \pi, is rotated through four right angles about the xx-axis. Find the exact volume of the solid of revolution.

By partsSolving equationVolume of revolutionStandard integrands中文
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Worked solution

8

1. Apply integration by parts to $C$ to prove the first relationship.

C=∫0πe−2xcos⁡x  dxC = \int_{0}^{\pi} \mathrm{e}^{-2x} \cos x \; \mathrm{d}x
Let u=e−2x  ⟹  dudx=−2e−2xu = \mathrm{e}^{-2x} \implies \frac{\mathrm{d}u}{\mathrm{d}x} = -2\mathrm{e}^{-2x}
Let dvdx=cos⁡x  ⟹  v=sin⁡x\frac{\mathrm{d}v}{\mathrm{d}x} = \cos x \implies v = \sin x
C=[e−2xsin⁡x]0π−∫0π−2e−2xsin⁡x  dxC = \left[ \mathrm{e}^{-2x} \sin x \right]_{0}^{\pi} - \int_{0}^{\pi} -2\mathrm{e}^{-2x} \sin x \; \mathrm{d}x
C=(e−2πsin⁡π−e0sin⁡0)+2∫0πe−2xsin⁡x  dxC = (\mathrm{e}^{-2\pi} \sin \pi - \mathrm{e}^{0} \sin 0) + 2 \int_{0}^{\pi} \mathrm{e}^{-2x} \sin x \; \mathrm{d}x
C=0+2S=2S(shown)C = 0 + 2S = 2S \quad \text{(shown)}

2. Apply integration by parts to $S$ to prove the second relationship.

S=∫0πe−2xsin⁡x  dxS = \int_{0}^{\pi} \mathrm{e}^{-2x} \sin x \; \mathrm{d}x
Let u=e−2x  ⟹  dudx=−2e−2xu = \mathrm{e}^{-2x} \implies \frac{\mathrm{d}u}{\mathrm{d}x} = -2\mathrm{e}^{-2x}
Let dvdx=sin⁡x  ⟹  v=−cos⁡x\frac{\mathrm{d}v}{\mathrm{d}x} = \sin x \implies v = -\cos x
S=[−e−2xcos⁡x]0π−∫0π2e−2xcos⁡x  dxS = \left[ -\mathrm{e}^{-2x} \cos x \right]_{0}^{\pi} - \int_{0}^{\pi} 2\mathrm{e}^{-2x} \cos x \; \mathrm{d}x
S=(−e−2πcos⁡π−(−e0cos⁡0))−2CS = \left(-\mathrm{e}^{-2\pi} \cos \pi - (-\mathrm{e}^{0} \cos 0)\right) - 2C
S=(−e−2π(−1)+1(1))−2CS = \left(-\mathrm{e}^{-2\pi}(-1) + 1(1)\right) - 2C
S=−2C+1+e−2π(shown)S = -2C + 1 + \mathrm{e}^{-2\pi} \quad \text{(shown)}

3. Solve the resulting simultaneous equations for the exact values of $C$ and $S$.

Substitute C=2SC = 2S into the second equation:
S=−2(2S)+1+e−2πS = -2(2S) + 1 + \mathrm{e}^{-2\pi}
S=−4S+1+e−2πS = -4S + 1 + \mathrm{e}^{-2\pi}
5S=1+e−2π  ⟹  S=1+e−2π55S = 1 + \mathrm{e}^{-2\pi} \implies S = \frac{1 + \mathrm{e}^{-2\pi}}{5}
Substitute SS back into C=2SC = 2S:
C=2(1+e−2π5)=2+2e−2π5C = 2\left(\frac{1 + \mathrm{e}^{-2\pi}}{5}\right) = \frac{2 + 2\mathrm{e}^{-2\pi}}{5}

4. Set up the integral for the volume of revolution and simplify using the double angle identity.

The volume VV rotated by four right angles (2π2\pi radians) is:
V=π∫0πy2  dxV = \pi \int_{0}^{\pi} y^2 \; \mathrm{d}x
V=π∫0π(e−xcos⁡(12x))2  dxV = \pi \int_{0}^{\pi} \left( \mathrm{e}^{-x} \cos\left(\frac{1}{2}x\right) \right)^2 \; \mathrm{d}x
V=π∫0πe−2xcos⁡2(12x)  dxV = \pi \int_{0}^{\pi} \mathrm{e}^{-2x} \cos^2\left(\frac{1}{2}x\right) \; \mathrm{d}x
Using the identity cos⁡2(12x)=1+cos⁡x2\cos^2\left(\frac{1}{2}x\right) = \frac{1 + \cos x}{2}:
V=π∫0πe−2x(1+cos⁡x2)  dxV = \pi \int_{0}^{\pi} \mathrm{e}^{-2x} \left( \frac{1 + \cos x}{2} \right) \; \mathrm{d}x
V=π2∫0πe−2x  dx+π2∫0πe−2xcos⁡x  dxV = \frac{\pi}{2} \int_{0}^{\pi} \mathrm{e}^{-2x} \; \mathrm{d}x + \frac{\pi}{2} \int_{0}^{\pi} \mathrm{e}^{-2x} \cos x \; \mathrm{d}x
V=π2∫0πe−2x  dx+π2CV = \frac{\pi}{2} \int_{0}^{\pi} \mathrm{e}^{-2x} \; \mathrm{d}x + \frac{\pi}{2} C

5. Evaluate the definite integral and calculate the exact volume.

Evaluate the first integral part:
∫0πe−2x  dx=[−12e−2x]0π=−12e−2π−(−12e0)=1−e−2π2\int_{0}^{\pi} \mathrm{e}^{-2x} \; \mathrm{d}x = \left[ -\frac{1}{2}\mathrm{e}^{-2x} \right]_{0}^{\pi} = -\frac{1}{2}\mathrm{e}^{-2\pi} - \left(-\frac{1}{2}\mathrm{e}^{0}\right) = \frac{1 - \mathrm{e}^{-2\pi}}{2}
Substitute this and CC into the volume expression:
V=π2(1−e−2π2)+π2(2+2e−2π5)V = \frac{\pi}{2} \left( \frac{1 - \mathrm{e}^{-2\pi}}{2} \right) + \frac{\pi}{2} \left( \frac{2 + 2\mathrm{e}^{-2\pi}}{5} \right)
V=π2(5(1−e−2π)+2(2+2e−2π)10)V = \frac{\pi}{2} \left( \frac{5(1 - \mathrm{e}^{-2\pi}) + 2(2 + 2\mathrm{e}^{-2\pi})}{10} \right)
V=π2(5−5e−2π+4+4e−2π10)V = \frac{\pi}{2} \left( \frac{5 - 5\mathrm{e}^{-2\pi} + 4 + 4\mathrm{e}^{-2\pi}}{10} \right)
V=π2(9−e−2π10)=π(9−e−2π)20V = \frac{\pi}{2} \left( \frac{9 - \mathrm{e}^{-2\pi}}{10} \right) = \frac{\pi(9 - \mathrm{e}^{-2\pi})}{20}

Answer

C=2+2e−2π5C = \frac{2 + 2\mathrm{e}^{-2\pi}}{5}, S=1+e−2π5S = \frac{1 + \mathrm{e}^{-2\pi}}{5}

Volume =π(9−e−2π)20= \frac{\pi(9 - \mathrm{e}^{-2\pi})}{20}

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