Respuesta :
In cylindrical coordinates,
[tex]\begin{cases}x(X,Y,Z)=X\cos Y\\y(X,Y,Z)=X\sin Y\\z(X,Y,Z)=Z\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=X\,\mathrm dX\,\mathrm dY\,\mathrm dZ[/tex]
Then the integral is
[tex]\displaystyle\iiint_Ex\,\mathrm dV=\int_{Y=0}^{Y=2\pi}\int_{X=4}^{X=6}\int_{Z=0}^{Z=X\cos Y+X\sin Y+10}X\cos Y\times X\,\mathrm dZ\,\mathrm dX\,\mathrm dY[/tex]
[tex]=\displaystyle\int_{Y=0}^{Y=2\pi}\int_{X=4}^{X=6}(10X^2\cos Y+X^3\cos^2Y+X^3\cos Y\sin Y)\,\mathrm dX\,\mathrm dY[/tex]
[tex]=\displaystyle\int_{Y=0}^{Y=2\pi}\left(\dfrac{1520}3\cos Y+260\cos^2Y+260\cos Y\sin Y\right)\,\mathrm dY=260\pi[/tex]
[tex]\begin{cases}x(X,Y,Z)=X\cos Y\\y(X,Y,Z)=X\sin Y\\z(X,Y,Z)=Z\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=X\,\mathrm dX\,\mathrm dY\,\mathrm dZ[/tex]
Then the integral is
[tex]\displaystyle\iiint_Ex\,\mathrm dV=\int_{Y=0}^{Y=2\pi}\int_{X=4}^{X=6}\int_{Z=0}^{Z=X\cos Y+X\sin Y+10}X\cos Y\times X\,\mathrm dZ\,\mathrm dX\,\mathrm dY[/tex]
[tex]=\displaystyle\int_{Y=0}^{Y=2\pi}\int_{X=4}^{X=6}(10X^2\cos Y+X^3\cos^2Y+X^3\cos Y\sin Y)\,\mathrm dX\,\mathrm dY[/tex]
[tex]=\displaystyle\int_{Y=0}^{Y=2\pi}\left(\dfrac{1520}3\cos Y+260\cos^2Y+260\cos Y\sin Y\right)\,\mathrm dY=260\pi[/tex]
The value of x dV is calculated by using the cylindrical coordinate and enclosed by z = 0 and z = x + y + 10 and by the cylinders x² + y² = 16 and x² + y² = 36 is 260π.
What is an area bounded by the curve?
When the two curves intersect then they bound the region is known as the area bounded by the curve.
Use cylindrical coordinates.
Evaluate x dv e ,
where e is enclosed by the planes z = 0 and z = x + y + 10 and by the cylinders x² + y² = 16 and x² + y² = 36.
We know that
[tex]\left\{\begin{matrix}x(X,Y,Z) &= X \cos Y \\y(X,Y,Z) &= Y \sin Y\\z(X,Y,Z) &= Z \end{matrix}\right[/tex]
Then we have
[tex]dV = dx\ dy\ dz = XdX \ dY\ dZ[/tex]
Then by the integration, we have
[tex]\begin{aligned} \int\int \int _{e} xdV &= \int_{Y=0}^{Y=2\pi}\int_{X=4}^{X=6}\int_{Z=0}^{Z= X\cos Y +X \sin Y + 10} X\cos Y \times XdXdYdZ \\\\ &= \int_{Y=0}^{Y=2 \pi}\int_{X=4}^{X=6} (10 X^{2}\cos Y + X^{3}\cos ^{2}Y+ X^{3} \cos Y \sin Y) dXdY \\\\ &= \int_{Y=0}^{Y=2\pi}\left ( \dfrac{1520}{3}\cos Y + 260 \cos ^{2}Y+260\cos Y \sin Y \right )dY\\\\ &=260 \pi \end{aligned}[/tex]
More about the area bounded by the curve link is given below.
https://brainly.com/question/24563834
