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[tex]\displaystyle\iiint_R\mathrm dV=\int_{y=-7}^{y=7}\int_{x=-\sqrt{49-y^2}}^{x=0}\int_{z=x}^{z=0}\mathrm dz\,\mathrm dx\,\mathrm dy[/tex]

Converting to cylindrical coordinates, the integral is equivalent to

[tex]\displaystyle\iiint_R\mathrm dV=\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}\int_{z=r\cos\theta}^{z=0}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_{\theta=\pi/2}^{\theta=3\pi/2}\int_{r=0}^{r=7}-r^2\cos\theta\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=-\displaystyle\left(\int_{\theta=\pi/2}^{3\pi/2}\cos\theta\,\mathrm d\theta\right)\left(\int_{r=0}^{r=7}r^2\,\mathrm dr\right)[/tex]
[tex]=\dfrac{2\times7^3}3=\dfrac{686}3[/tex]
okpalawalter8

We have that

Volume of the  wedge-shaped region

[tex]W=228.65[/tex]

Attached below is a picture of the wedge-shaped region contained in the cylinder

Bearing in mind this shape, we have that the coordinates identifying the cylinder are as follows

Co-ordinates

[tex]0 \leq r \leq 7,\frac{\pi}{2} \leq 0 \leq \frac{\pi}{2},0 \leq z \leq x[/tex]

Where

[tex]0 \leq z \leq x =rcos \phi[/tex]

Generally the equation for the volume of the  wedge-shaped region W  is mathematically given as

[tex]\int \int \int_W 1dv=\int^{7}_{0} \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \int^{rcos \phi}_{0}[/tex]

[tex]\int \int \int_W 1dv=\int^{7}_{0} r (\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}rcos\phi dv))dr[/tex]

[tex]\int \int \int_W 1dv=(\frac{(7)^3}{3}+\frac{(0)^3}{3})(sin {\pi/2}+sin{-\pi/2})[/tex]

[tex]\int \int \int_W 1dv=228.67[/tex]

Therefore

Volume of the  wedge-shaped region

[tex]W=228.65[/tex]

For more information on this please visit

https://brainly.com/question/16177070

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