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Question-219634




Question Number 219634 by Nicholas666 last updated on 29/Apr/25
Answered by MrGaster last updated on 01/May/25
  \begin{document}  \begin{align}  \int{x\geq 0,y\geq 1,0\leq z\leq 3-x+y}x\sin(\pi y)\,dx\,dy\,dz&=\int{0}^{\infty}\int{1}^{\infty}\int{0}^{3-x+y}x\sin(\pi y)\,dz\,dy\,dx\\  \int{0}^{3-x+y}x\sin(\pi y)\,dz&=x\sin(\pi y)(3-x+y)\\  \int{1}^{\infty}\int{0}^{\infty}x\sin(\pi y)(3-x+y)\,dx\,dy\\  \text{Consider the integration region:}3-x+y\geq 0&\Rightarrow y\geq x-3\text{,combined with}y\geq 1\\  \text{Divide the region into two parts:}\\    1. &\0\leq x\leq 4,\y\geq 1\\    2. &\x>4,\y\geq x-3\\  \text{First part:}\int{0}^{4}\int{1}^{\infty}x\sin(\pi y)(3-x+y)\,dy\,dx\\  \text{The integral with respect to}y\text{diverges,so consider the finite region:}1\leq x\leq 3,\1\leq y\leq x\\  \int{1}^{3}\int{1}^{x}x\sin(\pi y)(3-x+y)\,dy\,dx\\  \int{1}^{x}\sin(\pi y)(3-x+y)\,dy&=\frac{-3\cos(\pi x)+x-2}{\pi}\\  \int{1}^{3}x\left(\frac{-3\cos(\pi x)+x-2}{\pi}\right)\,dx&=\frac{2}{3\pi}\\  \boxed{\dfrac{2}{3\pi}}  \end{align}  \end{document}  ” ></figure>
</div>
<div style= $$ \\ $$\begin{document}
\begin{align}
\int{x\geq 0,y\geq 1,0\leq z\leq 3-x+y}x\sin(\pi y)\,dx\,dy\,dz&=\int{0}^{\infty}\int{1}^{\infty}\int{0}^{3-x+y}x\sin(\pi y)\,dz\,dy\,dx\\
\int{0}^{3-x+y}x\sin(\pi y)\,dz&=x\sin(\pi y)(3-x+y)\\
\int{1}^{\infty}\int{0}^{\infty}x\sin(\pi y)(3-x+y)\,dx\,dy\\
\text{Consider the integration region:}3-x+y\geq 0&\Rightarrow y\geq x-3\text{,combined with}y\geq 1\\
\text{Divide the region into two parts:}\\

1. &\0\leq x\leq 4,\y\geq 1\\

2. &\x>4,\y\geq x-3\\
\text{First part:}\int{0}^{4}\int{1}^{\infty}x\sin(\pi y)(3-x+y)\,dy\,dx\\
\text{The integral with respect to}y\text{diverges,so consider the finite region:}1\leq x\leq 3,\1\leq y\leq x\\
\int{1}^{3}\int{1}^{x}x\sin(\pi y)(3-x+y)\,dy\,dx\\
\int{1}^{x}\sin(\pi y)(3-x+y)\,dy&=\frac{-3\cos(\pi x)+x-2}{\pi}\\
\int{1}^{3}x\left(\frac{-3\cos(\pi x)+x-2}{\pi}\right)\,dx&=\frac{2}{3\pi}\\
\boxed{\dfrac{2}{3\pi}}
\end{align}
\end{document}

Commented by MrGaster last updated on 01/May/25
Complete:\begin{document} \begin{align} \int{x\geq 0,y\geq 1,0\leq z\leq 3-x+y}x\sin(\pi y)\,dx\,dy\,dz&=\int{0}^{\infty}\int{1}^{\infty}\int{0}^{3-x+y}x\sin(\pi y)\,dz\,dy\,dx\\ \int{0}^{3-x+y}x\sin(\pi y)\,dz&=x\sin(\pi y)(3-x+y)\\ \int{1}^{\infty}\int{0}^{\infty}x\sin(\pi y)(3-x+y)\,dx\,dy\\ \text{Consider the integration region:}3-x+y\geq 0&\Rightarrow y\geq x-3\text{,combined with}y\geq 1\\ \text{Divide the region into two parts:}\\ 1. &\0\leq x\leq 4,\y\geq 1\\ 2. &\x>4,\y\geq x-3\\ \text{First part:}\int{0}^{4}\int{1}^{\infty}x\sin(\pi y)(3-x+y)\,dy\,dx\\ \text{The integral with respect to}y\text{diverges,so consider the finite region:}1\leq x\leq 3,\1\leq y\leq x\\ \int{1}^{3}\int{1}^{x}x\sin(\pi y)(3-x+y)\,dy\,dx\\ \int{1}^{x}\sin(\pi y)(3-x+y)\,dy&=\frac{-3\cos(\pi x)+x-2}{\pi}\\ \int{1}^{3}x\left(\frac{-3\cos(\pi x)+x-2}{\pi}\right)\,dx&=\frac{2}{3\pi}\\ \boxed{\dfrac{2}{3\pi}} \end{align} \end{document}
Commented by Nicholas666 last updated on 02/May/25
false
$${false} \\ $$

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