91Ó°ÊÓ

To find the bounds for \(x\) and \(y\), we need to find the intersection of the cylinder with the \(xy\)-plane. This can be found by setting \(z=0\) in both plane equations and solving for \(x\) and \(y\). For the plane \(z = 3-x\), we have: \(0 = 3-x\) \(x = 3\) For the plane \(z = x-3\), we have: \(0 = x-3\) \(x = 3\) Since both planes intersect at \(x=3\), we can now find the range of \(y\) by substituting this into the cylinder equation: \(x^{2} + 4y^{2} = 4\) \(3^{2} + 4y^{2} = 4\) \(9 + 4y^{2} = 4\) \(4y^{2} = -5\) Since this equation has no solution for \(y\), we conclude that the cylinder does not intersect the \(xy\)-plane. However, we can use the cylinder equation directly to find the bounds for \(x\) and \(y\): \(x^{2} + 4y^{2} = 4\) Let's first solve for x: \(x^2 = 4 - 4y^2\) \(-x^2 = 4y^2 - 4\) \(x^2 = 4(1 - y^2)\) \(x = \pm 2\sqrt{1 - y^2}\) Now let's solve for y: \(4y^2 = 4 - x^2\) \(y^2 = 1 - \frac{x^2}{4}\) \(y = \pm \frac{1}{2}\sqrt{4 - x^2}\) Therefore, our bounds for \(x\) and \(y\) are: \(-2\sqrt{1-y^2} \leq x \leq 2\sqrt{1-y^2}\) and \(-\frac{1}{2}\sqrt{4-x^2} \leq y \leq \frac{1}{2}\sqrt{4-x^2}.\)

Using the two planes defining the wedge, we will find the range of integration for \(z\): Lower bound for \(z\) (plane \(z=3-x\)): \(z \geq 3-x\) Upper bound for \(z\) (plane \(z=x-3\)): \(z \leq x-3\)

Using the bounds of integration obtained in Steps 1 and 2, we can now set up the triple integral for the volume: \(\displaystyle\int\limits_{-1}^{1} \int\limits_{-\frac{1}{2}\sqrt{4 - 4y^2}}^{\frac{1}{2}\sqrt{4-4y^2}} \int\limits_{3-x}^{x-3} dz\, dx\, dy\)

Now let's evaluate the triple integral: \(\displaystyle\int\limits_{-1}^{1} \int\limits_{-\frac{1}{2}\sqrt{4-4y^2}}^{\frac{1}{2}\sqrt{4-4y^2}} [(x-3)-(3-x)]\, dx\, dy\) Simplifying the integrand: \(\displaystyle\int\limits_{-1}^{1} \int\limits_{-\frac{1}{2}\sqrt{4-4y^2}}^{\frac{1}{2}\sqrt{4-4y^2}} 2x-6\,dx\,dy\) First, evaluate the inner integral w.r.t. x: \(\left[\mathrm{x^2\, - 6x}\right]_{-\frac{1}{2}\sqrt{4-4y^2}}^{\frac{1}{2}\sqrt{4-4y^2}}\) Now evaluate the \(\displaystyle\int\limits_{-1}^{1} [4(1-y^2)-6\sqrt{1-y^2}]\,dy\) Using substitution, let \(y = \sin{u}\). Then, \(dy = \cos{u} du\) and the integral becomes: \(\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [4(1-\sin^2{u})-6\sqrt{1-\sin^2{u}}]\cos{u}\,du\) Evaluate the integral: \(4\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^2{u})\,du - 6\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^2{u} - \cos^2{u})\cos{u}\,du\) The first integral evaluates to \(\frac{4\pi}{2}\) and the second integral is 0 (since the integrand is now 0): The volume of the wedge is \(\frac{4\pi}{2}\) or \(2\pi\) which is the final answer. The volume of the wedge created by the planes \(z=3-x\) and \(z=x-3\) and the cylinder \(x^{2}+4y^{2}=4\) is \(2\pi\).