Evaluate the iterated integral. integral.
0^pi integral_0^1 integral_0^Squareroot 1 - z^2 z sin(x) dy dz dx
_______________

Answers

Answer 1

The given iterated integral is [tex]\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx[/tex]. The value of this integral is evaluated as 2/3.

The iterated integral is obtained by repeatedly integrating multivariate functions with a single variable each time. The given iterated integral is [tex]\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx[/tex]. This integral is evaluated in the following order [tex]\int dy[/tex], [tex]\int dz[/tex], and [tex]\int dx[/tex].

First, integrate the above expression with respect to y. Then, substitute the limit values, and we get,

[tex]\begin{aligned}\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx&=\int\limits^{\pi}_b\int\limits^1_0z\sin(x)\left[\int\limits^{\sqrt{1-z^2}}_0 \;dy\right]\;dz \;dx\\&=\int\limits^{\pi}_b\int\limits^1_0z\sin(x)[y]^{\sqrt{1-z^2}}_{0}\;dz \;dx\\&=\int\limits^{\pi}_b\int\limits^1_0z\sin(x)\left(\sqrt{1-z^2}\right)\;dz \;dx\end{aligned}[/tex]

Now, integrating with respect to z, we get,

[tex]\begin{aligned}\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx&=\int\limits^{\pi}_b\int\limits^1_0\sin(x)\;z\left(\sqrt{1-z^2}\right)\;dz \;dx\\&=-\frac{1}{2}\int\limits^{\pi}_b\sin x\left[\int\limits^1_0-2z\left(\sqrt{1-z^2}\right)\;dz\right]\;dx\end{aligned}[/tex]

Substitute, 1-z²=u and -2zdz=du in the above expression, and we get,

[tex]\begin{aligned}\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx&=-\frac{1}{2}\int\limits^{\pi}_b\sin x\left[\int\limits^0_1\left(\sqrt{u^2}\right)\;du\right]\;dx\\&=-\frac{1}{2}\int\limits^{\pi}_b\sin x\left[\frac {u^{3/2}}{3/2}\right]^0_1\;dx\\&=\frac{1}{3}\int\limits^{\pi}_b\sin x\left[u^{3/2}\right]^1_0\;dx\\&=\frac{1}{3}\int\limits^{\pi}_b\sin x\;dx\end{aligned}[/tex]

Finally, integrating with respect to x and we get,

[tex]\begin{aligned}\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx&=\frac{1}{3}\int\limits^{\pi}_b\sin x\;dx\\&=\frac{1}{3}[-(-1)-(-(1))]\\&=\frac{1}{3}[2]\\&=\frac{2}{3}\end{aligned}[/tex]

The required answer for the given integral is 2/3.

The complete question is -

Evaluate the iterated integral [tex]\int\limits^{\pi}_b\int\limits^1_0\int\limits^{\sqrt{1-z^2}}_0 z\sin(x)\;dy\;dz \;dx[/tex].

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Step-by-step explanation:

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[tex]\Large\boxed{\tt Anwer:~80+30h}[/tex]

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Answers

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