Find the volume of the solid enclosed by the paraboloids $z=16-3x^2-3y^2$ and $z=4$ so what i did is this $4=16-3x^2-3y^2$ and I'm not sure about the following steps.
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By setting the two given constraints to be true, all you are doing is asserting that the point $(x, y, z)$ exists in the intersection between the two regions. Think about what you are doing, and how it differs from what the question is asking. – Myridium Jul 22 '14 at 06:17
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$4 = 16 - 3x^2 - 3y^2 \to x^2 + y^2 = 2^2$, this tells you about the base disk that you integrate on. Using cylindrical coordinates: $x = r\cos\theta$, $y = r\sin\theta$, $z = z$, we have: $V = \displaystyle \int_{0}^{2\pi}\int_{0}^2\int_{4}^{16-3r^2} rdzdrd\theta$
DeepSea
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Region in the plane $xy$: $$ 4 = 16 - 3x^2 - 3y^2 \quad \Rightarrow \quad x^2 + y^2 = 4 \quad \Rightarrow \quad r = 2 $$ Using polar coordinates, we have $$ V = \int_R\int f(x,y)dA = \int_{0}^{2\pi}\int_{0}^{2}(16 - 3r^2)rdrd\theta = 2\pi \int_{0}^{2}(16r - 3r^2)dr = 48\pi $$
Mathsource
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As you see, I just wanted to illustrate what other post tried to indicate mathematically:
Mikasa
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