Problem 28 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \int_{0}^{12-2 y-3 z} \frac{1}{y} d x d z d y$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \int_{0}^{12-2 y-3 z} \frac{1}{y} d x d z d y$$ Answer: The value of the triple integral is $$10 + 24\ln|6|$$

Step 1: Integrate with respect to x

We will first integrate the function with respect to x. The integral will be: $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \left[\frac{1}{y} \int_{0}^{12-2 y-3 z} dx \right] d z d y$$ Integrating with respect to x gives us: $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \frac{1}{y}(12-2 y-3 z) \Big{|}_0^{12-2 y-3 z} d z d y = \int_{1}^{6} \int_{0}^{4-2 y / 3} \frac{(12-2 y-3 z)}{y} d z d y$$

Step 2: Integrate with respect to z

Now, we will integrate the resulting function with respect to z $$\int_{1}^{6} \left[\frac{1}{y} \int_{0}^{4-2 y / 3} (12-2 y-3 z) d z \right] d y$$ We integrate with respect to z: $$\int_{1}^{6} \left[\frac{1}{y} \left(12z-2 yz-\frac{3}{2}z^2\right)\Big{|}_0^{4-2 y / 3} \right] d y$$

Step 3: Substitute the integration bounds for z

Now, we substitute the integration bounds for z to obtain $$\int_{1}^{6} \left[\frac{1}{y} \left(12(4-\frac{2y}{3})-2y(4-\frac{2y}{3})-\frac{3}{2}(4-\frac{2y}{3})^2\right) \right] d y$$ Let's simplify the expression inside the integral $$\int_{1}^{6} \left[\frac{48 - 8y + \frac{4y^2}{3} - 8y + \frac{4y^2}{3} +\frac{1}{2}(16 - \frac{32y}{3}+ \frac{4y^2}{3}) }{y} \right] d y$$ $$\int_{1}^{6} \left[\frac{1}{2y} (48 +\frac{20y^2}{3} -32y) \right] d y$$

Step 4: Integrate with respect to y

Finally, we integrate the simplified expression with respect to y: $$\int_{1}^{6} \left[\frac{1}{2} (48\ln(y) +\frac{20y}{3} -32) \right] d y$$ Now, we evaluate the integral with respect to y $$\frac12\left[\left(\frac{20y^2}{6}-32y+48\ln|y|\right)\right]_{1}^{6}$$

Step 5: Substitute the integration bounds for y and find the result

Now we substitute the bounds for y to find the final result $$=\frac12\left[\left(\frac{20\cdot6^2}{6}-32\cdot6+48\ln|6|\right)-\left(\frac{20\cdot1^2}{6}-32\cdot1+48\ln|1|\right)\right]$$ Evaluating the expression and simplifying $$=\frac12\left(240-192+48\ln|6|-(\frac{20}{6}-32) \right)=\frac12\left(20+48\ln|6|\right)$$ So the final result is $$\int_{1}^{6} \int_{0}^{4-2 y / 3} \int_{0}^{12-2 y-3 z} \frac{1}{y} d x d z d y = 10 + 24\ln|6|$$