91Ó°ÊÓ

We first need to find where these two paraboloids intersect each other. Let's set the z-coordinates of the two equations equal to each other and solve for x and y: $$2x^2 + y^2 = 27 - x^2 - 2y^2$$ Now, we will rewrite this equation as a single equation in x and y coordinate, which represents the intersection curves of the paraboloids.

Let's rewrite the equation as follows: $$3x^2 + 3y^2 = 27$$ Divide the equation by 3: $$x^2 + y^2 = 9$$ The intersection curve is a circle with a center at the origin and radius of 3 in the xy-plane.

Now, let's find the volume bounds. Our goal is to integrate over the region enclosed by these two surfaces along the z-axis. We can parameterize this region by cylindrical coordinates: $$x = r\cos(\theta)$$ $$y = r\sin(\theta)$$ Since the intersection curve is a circle of radius 3, the bounds for r and θ are: $$0 \leq r \leq 3$$ $$0 \leq \theta \leq 2\pi$$ For the z bounds, we will find the minimum and maximum z-coordinate of the solid at a given point (r, θ). We can express the two surfaces in cylindrical coordinates to find the minimum and maximum z values: 1) \(z = 2r^2\cos^2(\theta) + r^2\sin^2(\theta)\) 2) \(z = 27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta)\)

Now we need to set up the triple integral which will find the volume of the solid. We can write the integral with respect to dz, dr, and dθ in the following order: $$V = \int_{0}^{2\pi} \int_{0}^{3} \int_{2r^2\cos^2(\theta) + r^2\sin^2(\theta)}^{27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta)} rdzdrdθ$$

Now we can calculate the triple integral: $$V = \int_{0}^{2\pi} \int_{0}^{3} [27 - r^2\cos^2(\theta) - 2r^2\sin^2(\theta) - (2r^2\cos^2(\theta) + r^2\sin^2(\theta))]r dr d\theta$$ Simplify the integrand: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3r^2\cos^2(\theta) - 3r^2\sin^2(\theta))r dr d\theta$$ We can factor out the 3 from the integrand: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3(r^2\cos^2(\theta) + r^2\sin^2(\theta)))r dr d\theta$$ Now, we can notice that \(\cos^2(\theta) + \sin^2(\theta) = 1\). So, the integrand simplifies to: $$V = \int_{0}^{2\pi} \int_{0}^{3} (27 - 3r^2)r dr d\theta$$ Now, let's calculate the integral: $$V = \int_{0}^{2\pi} \left[\int_{0}^{3} (27r - 3r^3) dr \right] d\theta$$ Integrate with respect to r: $$V = \int_{0}^{2\pi} \left[\frac{27}{2}r^2 - \frac{3}{4}r^4 \right]_{0}^{3} d\theta$$ Substitute r = 3 and subtract the result of r = 0: $$V = \int_{0}^{2\pi} [27(3)^2/2 - 3(3)^4/4] d\theta$$ Simplify: $$V = \int_{0}^{2\pi} \left[\frac{81}{2} - \frac{81}{4} \right] d\theta$$ Calculate the integral with respect to θ: $$V = \left[\frac{81}{4}\theta \right]_{0}^{2\pi}$$ Substitute θ = 2π and subtract the result of θ = 0: $$V = \frac{81}{4}(2\pi)$$ Thus, the volume of the solid bounded by the two paraboloids is: $$V = \frac{81}{2}\pi$$