The volume of the region in first octant is 9.
Given that the region in the first octant bounded by the cylinder r=3 and the plane z=y.
The graph is shown below.
we are given that first octant is bound by r=3 and z=y
We will use the conversion formula, i.e.
x²+y²=r² where x=rcos∅ and y=rsin∅
Firstly, we will find bound octant
0≤∅≤π/2
0≤r≤3
0≤z≤r sin∅
Now, we can set up integral
[tex]V=\int_{0}^{\frac{\pi}{2}}\int_{0}^{3}\int_{0}^{rsin(\theta)}rdzdrd\theta[/tex]
Further, we can solve it.
Firstly, we solve integral for z then we solve for r and after that we will solve for ∅, we get
[tex]\begin{aligned}V&=\int_{0}^{\frac{\pi}{2}}\int_{0}^{3}r^2 \sin (\theta)dr d\theta\\&=\int_{0}^{\frac{\pi}{2}}\frac{1}{3}r^3\sin \theta d\theta\left|_{0}^{3}\\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{3}\sin\theta d\theta(27-0)\right)\\ &=\int_{0}^{\frac{\pi}{2}}9\sin \theta d\theta\\ &=0-(-9)\\ &=9\end[/tex]
Hence, the volume of the region in the first octant is bounded by the cylinder r=3 and the plane z=y is 9.
Learn more about volume from here brainly.com/question/4036200
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