Día: 3 de diciembre de 2025

Nos piden calcular \[\iiint_Q\ dV=\int_{0}^{1}\int_{1}^{2}\int_{0}^{2} \ dz\,dy\,dx.\] Aplicando la definición: \[\int_{0}^{1}\int_{1}^{2}\int_{0}^{2} \ dz\,dy\,dx=\int_{0}^{1}\left[\int_{1}^{2}\left[z\right|_{0}^{2}\,dy\right]\,dx=\] \[=\int_{0}^{1}2\left[y\right|_{1}^{2}\,dx=2\left[x\right|_{0}^{2}=4\]

Nos piden calcular \[\iiint_Q\ dV=\int_{0}^{1}\int_{0}^{1}\int_{0}^{x-y^2} \ dz\,dy\,dx.\] Veámoslo: \[\int_{0}^{1}\int_{0}^{1}\int_{0}^{x-y^2} \ dz\,dy\,dx=\int_{0}^{1}\int_{0}^{1}\left[\int_{0}^{x-y^2} \ dz\right]\,dy\,dx=\]\[=\int_{0}^{1}\left[\int_{0}^{1}\left[z\right|_{0}^{x-y^2}\,dy\right]\,dx=\int_{0}^{1}\left[\int_{0}^{1}\left[{x-y^2}\right]\,dy\right]\,dx=\] \[=\int_{0}^{1}\left[yx-\frac{y^3}{3}\right|_0^1\,dx=\int_{0}^{1}\left[x-\frac{1}{3}\right]\,dx=\left[\frac{x^2}{2}-\frac{x}{3}\right|_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\]

Resolvamos primero
\[ I_1 = \int_{0}^{x} \cos(x+y+z)dz= \left[ \sin(x+y+z) \right]_{z=0}^{z=x}=\sin(2x+y) – \sin(x+y) \]
Ahora,
\[
\begin{align}
I_2 &= \int_{0}^{y} \left( \sin(2x+y) – \sin(x+y) \right) dx \\
&= \left[ -\frac{1}{2}\cos(2x+y) + \cos(x+y) \right]_{x=0}^{x=y}\\
&=-\frac{1}{2}\cos(3y) + \cos(2y)-\frac{1}{2}\cos(y)
\end{align}
\]
Por último,
\[
\begin{align}
I &= \int_{0}^{1} \left( \cos(2y) – \frac{1}{2}\cos(3y) – \frac{1}{2}\cos(y) \right) dy\\
&= \left[ \frac{1}{2}\sin(2y) – \frac{1}{6}\sin(3y) – \frac{1}{2}\sin(y) \right]_{0}^{1}\\
&= \frac{1}{2}\sin(2) – \frac{1}{6}\sin(3) – \frac{1}{2}\sin(1)
\end{align}
\]

\[
\begin{align}
I &=\int_{0}^{\sqrt{\pi}}\int_{0}^{x}\int_{0}^{xz} x^2\, \sin(y)dy\,dz\,dx \\
&=\int_{0}^{\sqrt{\pi}}\int_{0}^{x}\left(\int_{0}^{xz} x^2\, \sin(y)dy\right)\,dz\,dx \\
&=\int_{0}^{\sqrt{\pi}}\int_{0}^{x}\left(x^2 \left[ -\cos(y) \right]_{y=0}^{y=xz}\right)\,dz\,dx \\
&=\int_{0}^{\sqrt{\pi}}\int_{0}^{x}\left(x^2 \left( 1 – \cos(xz) \right)\right)\,dz\,dx \\
&=\int_{0}^{\sqrt{\pi}}\left(\int_{0}^{x}\left(x^2 \left( 1 – \cos(xz) \right)\right)\,dz\right)\,dx \\
&=\int_{0}^{\sqrt{\pi}}\left(x^2 \left[ z – \frac{1}{x}\sin(xz) \right]_{z=0}^{z=x}\right)\,dx \\
&=\int_{0}^{\sqrt{\pi}}\left(x^3 -x\sin(x^2)\right)\,dx \\
&\mbox{(hacemos un cambio de variable}\,u=x^2 \mbox{para resolver}\, \int \,x\sin(x^2)\,dx\\
&=\left[ \frac{x^4}{4} \right]_{0}^{\sqrt{\pi}}+ \frac{1}{2} \left[ -\cos(u) \right]_{0}^{\pi}\\
&=\frac{\pi^2}{4} – 1
\end{align}
\]

Vamos a calcularla paso a paso.

Paso 1: Integral interior (respecto a \(x\))

Integramos \(\left(\frac{z}{y+1}\right)\) con respecto a \(x\):
\[
\begin{aligned}
I_1 &= \int_{0}^{\sqrt{1-z^2}} \left(\frac{z}{y+1}\right)dx \\
&= \left(\frac{z}{y+1}\right) [x]_{x=0}^{x=\sqrt{1-z^2}} \\
&= \left(\frac{z}{y+1}\right) (\sqrt{1-z^2} – 0) \\
&= \frac{z\sqrt{1-z^2}}{y+1}
\end{aligned}
\]

Paso 2: Integral media (respecto a \(z\))

Sustituimos e integramos con respecto a \(z\):
\[
\begin{aligned}
I_2 &= \int_{0}^{1} \frac{z\sqrt{1-z^2}}{y+1} dz \\
&= \frac{1}{y+1} \int_{0}^{1} z\sqrt{1-z^2} dz
\end{aligned}
\]
Para la integral interna \(\int_{0}^{1} z\sqrt{1-z^2} dz\), usamos la sustitución \(u = 1 – z^2\), de donde \(z\,dz = -\frac{1}{2}du\).

\[
\begin{aligned}
\int_{0}^{1} z\sqrt{1-z^2} dz &= \int_{1}^{0} \sqrt{u} \left(-\frac{1}{2}du\right) \\
&= \frac{1}{2} \int_{0}^{1} u^{1/2} du \\
&= \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} \\
&= \frac{1}{3} (1^{3/2} – 0^{3/2}) = \frac{1}{3}
\end{aligned}
\]
Sustituyendo el resultado en \(I_2\):
\[
I_2 = \frac{1}{y+1} \cdot \frac{1}{3} = \frac{1}{3(y+1)}
\]

Paso 3: Integral exterior (respecto a \(y\))

Resolvemos la última integral con respecto a \(y\):
\[
\begin{aligned}
I &= \int_{0}^{1} \frac{1}{3(y+1)} dy \\
&= \frac{1}{3} \int_{0}^{1} \frac{1}{y+1} dy \\
&= \frac{1}{3} \left[ \ln|y+1| \right]_{0}^{1} \\
&= \frac{1}{3} \left( \ln(2) – \ln(1) \right) \\
&= \frac{1}{3} \ln(2)
\end{aligned}
\]

A.)