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in Definite Integrals by (55 points)
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Find the moment of inertia of the area under the curve y=sin x from x=0 to x=π with respect to y-axis.

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1 Answer

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Iy = \(\iint\limits_{R}\mathrm x^2dA\)

\(=\int\limits^\pi_0\left(\int^{\sin \mathrm x}_0\mathrm x^2dy\right)d\mathrm x\)

\(=\int\limits^\pi_0\mathrm x^2(y)^{\sin \mathrm x}_0 d\mathrm x\)

\(=\int\limits^{\pi}_0 \mathrm x^2\sin \mathrm x \,d\mathrm x\)

\(=\left[\mathrm x^2\int \sin \mathrm x\,d\mathrm x - \int\left(2\mathrm x\int \sin \mathrm x d\mathrm x\right)d\mathrm x\right]^\pi_0\)

\(=\left[-\mathrm x^2\cos\mathrm x+\int 2\mathrm x\cos\mathrm x\,d\mathrm x\right]^\pi_0\)

\(=[-\mathrm x^2\cos\mathrm x+2\mathrm x\)\(\int \cos\mathrm x\,d\mathrm x-\int 2(\int \cos\mathrm x \,d\mathrm x)d\mathrm x]^\pi_0\)

\(=[-\mathrm x^2\cos\mathrm x+2\mathrm x\sin \mathrm x-\int 2\sin \mathrm x d\mathrm x]^\pi_0\)

\(=[-\mathrm x^2\cos\mathrm x+2\mathrm x \sin \mathrm x+2\cos\mathrm x]^\pi_0\)

\(=(-\pi^2\cos\pi +2\pi \sin \pi+2\cos\pi)\) \(-(2\cos 0)\)

\(=\pi^2-2-2\)        \((\because \cos \pi =\lambda, \cos 0 =1\) & \(\sin \pi =0)\)

\(=\pi^2-4\)

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