Problem 40 - Calculus Early Transcendentals

Question

Rewrite the following integrals using the indicated order of integration and then evaluate the resulting integral. $$\int_{0}^{1} \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} d z d y d x \text { in the order } d y d z d x$$

Step-by-Step Solution

Answer

Based on the given problem and step-by-step solution, create the following short answer: Question: Rewrite the integral $$\int_{0}^{1} \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} d z d y d x$$ in the order dy dz dx and find its value. Answer: The rewritten integral is $$\int_{0}^{1} \int_{0}^{\sqrt{4-y^{2}}} \int_{-2}^{2} d y d z d x$$ and its value is $$\frac{4}{3}(9 \sqrt{3} - 32)$$

Step 1: Analyze the original limits of integration

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: \(0 \leq x \leq 1\)

Step 2: Sketch the region of integration

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

Step 3: Determine the new limits of integration for dy dz dx order

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: $$\int_{0}^{1} \int_{0}^{\sqrt{4-y^{2}}} \int_{-2}^{2} d y d z d x$$

Step 4: Evaluate the resulting integral

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: $$\int_{0}^{1} \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} d z d y d x = \frac{4}{3}(9 \sqrt{3} - 32)$$