202109161118 Exercises 3.3.A (Q3)

For Exercises 1-8, evaluate the given triple integral.

$\displaystyle{\int_{0}^{\pi}\int_{0}^{x}\int_{0}^{xy}x^2\sin z\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x}$

Solution.

$\begin{aligned} &\qquad \int_{0}^{\pi}\int_{0}^{x}\int_{0}^{xy}x^2\sin z\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi}\int_{0}^{x}\big( -x^2\cos z\big)\big|_{z=0}^{z=xy}\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi}\int_{0}^{x}(-x^2)(\cos xy-1)\,\mathrm{d}y\,\mathrm{d}x \\ & = \int_{0}^{\pi} (-x^2)\bigg(\frac{\sin xy}{x}-y\bigg)\bigg|_{y=0}^{y=x}\,\mathrm{d}x \\ & = \int_{0}^{\pi} (-x^2)\bigg( \frac{\sin x^2}{x}-x \bigg)\,\mathrm{d}x \\ & = \int_{0}^{\pi} (x^3 - x\sin x^2)\,\mathrm{d}x \\ & = \int_{0}^{\pi}x^3\,\mathrm{d}x - \int_{0}^{\pi} x\sin x^2\,\mathrm{d}x \\ & = \bigg( \frac{x^4}{4} \bigg)\bigg|_{0}^{\pi} - \bigg[ \cdots \bigg] \\ & = \enspace \dots \textrm{ (to be continued) }\dots \end{aligned}$

Roughwork. (Integration by parts)

$\begin{aligned} & \qquad \int x\sin x^2\,\mathrm{d}x \\ & = \int \sin x^2\,\mathrm{d}\bigg(\frac{x^2}{2}\bigg) \\ & = \frac{1}{2}\cdot\int \sin x^2\,\mathrm{d}(x^2) \\ & = - \frac{\cos x^2}{2} \\ \end{aligned}$

$\begin{aligned} & = \dots \textrm{ (to continue) }\dots \\ & = \bigg(\frac{x^4}{4}\bigg)\bigg|_{0}^{\pi} - \bigg( -\frac{\cos x^2}{2} \bigg)\bigg|_{0}^{\pi} \\ & = \bigg(\frac{\pi^4}{4}\bigg) + \bigg(\frac{\cos\pi^2}{2} - \frac{1}{2}\bigg) \\ & = -\frac{1}{2} + \frac{\pi^4}{4} + \frac{\cos\pi^2}{2} \\ \end{aligned}$

M	T	W	T	F	S	S
		1	2	3	4	5
6	7	8	9	10	11	12
13	14	15	16	17	18	19
20	21	22	23	24	25	26
27	28	29	30

Leave a Reply Cancel reply