91Ó°ÊÓ

In the given triple integral, we have integration in the order dz dy dx with the following limits: - For z: \(0 \leq z \leq \sqrt{4 - y^2}\) - For y: \(-2 \leq y \leq 2\) - For x: \(0 \leq x \leq 1\)

Sketching the region of integration can help us understand and visualize the given triple integral. The region can be defined as a solid object with the following bounds: - x-axis: \(0 \leq x \leq 1\) - y-axis: \(-2 \leq y \leq 2\) - z-axis: \(0 \leq z \leq \sqrt{4 - y^2}\), This is a semi-circle with a radius of 2 on the xz-plane

We want to rewrite the given integral using the order dy dz dx. The new limits can be determined using the region of integration. - For y: As the semi-circle lies between \(-2 \leq y \leq 2\), y limits remain the same. - For z: Since z is bounded by the semicircle above and the xz-plane below, we have \(0 \leq z \leq \sqrt{4 - y^2}\). - For x: x is bounded between the values \(0 \leq x \leq 1\). The new integral becomes: $$\int_{0}^{1} \int_{0}^{\sqrt{4-y^{2}}} \int_{-2}^{2} d y d z d x$$

We'll evaluate the integral in the order given using the new arrangement and limits: $$\begin{aligned} \int_{0}^{1} \int_{0}^{\sqrt{4-y^{2}}} \int_{-2}^{2} d y d z d x &= \int_{0}^{1} \int_{0}^{\sqrt{4-y^{2}}} \Big[\int_{-2}^{2} dy\Big] dz dx \\ &= \int_{0}^{1}\int_{0}^{\sqrt{4-y^{2}}} (y\Big|_{-2}^{2}) dz dx \\ &= 4\int_{0}^{1}\int_{0}^{\sqrt{4-y^{2}}} dz dx \\ &= 4\int_{0}^{1} (z\Big|_{0}^{\sqrt{4-y^{2}}}) dx \\ &= 4\int_{0}^{1}(\sqrt{4-y^{2}}) dx \end{aligned}$$ Now, we can finish evaluating the integral: $$\begin{aligned} 4\int_{0}^{1}(\sqrt{4-y^{2}}) dx &= 4\Big[\frac{1}{3}(4-y^2)^{\frac{3}{2}}\Big|_{0}^{1}\Big] \\ &= 4\Big[\frac{1}{3}(4-1)^{\frac{3}{2}} - \frac{1}{3}(4-0)^{\frac{3}{2}}\Big] \\ &= 4\Big[\frac{1}{3}(3)^{\frac{3}{2}} - \frac{1}{3}(4)^{\frac{3}{2}}\Big] \\ &= \frac{4}{3}(9 \sqrt{3} - 32) \end{aligned}$$ So, the resulting integral is: $$\int_{0}^{1} \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} d z d y d x = \frac{4}{3}(9 \sqrt{3} - 32)$$