Problem 9 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the triple integral \(\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y\) Answer: The triple integral evaluates to \(0\).

Step 1: Inner integral for z

Evaluate the inner integral for the \(z\) variable: $$\int_{1}^{e} \frac{x y^{2}}{z} d z$$ Since both \(x\) and \(y\) are constants with respect to \(z\), they can be factored out of the integral: $$x y^{2} \int_{1}^{e} \frac{1}{z} d z$$ The antiderivative of \(\frac{1}{z}\) is \(\ln |z|\). Evaluate it at the given limits: $$x y^{2}(\ln |e| - \ln |1|)$$ Since \(\ln |e| = 1\) and \(\ln |1| = 0\), we get $$x y^{2}(1 - 0) = x y^{2}$$

Step 2: Middle integral for x

Now, we will take the result from Step 1 and evaluate the middle integral for the \(x\) variable: $$\int_{-2}^{2} x y^{2} d x$$ The antiderivative of \(x y^{2}\) with respect to \(x\) is \(\frac{1}{2} xy^2\). Evaluate it at the given limits: $$\frac{1}{2} y^{2} (2^2 - (-2)^2)$$ This simplifies to: $$\frac{1}{2} y^{2} (4 - 4) = 0$$

Step 3: Outer integral for y

Finally, we will take the result from Step 2 and evaluate the outer integral for the \(y\) variable: $$\int_{1}^{2} 0 d y$$ The antiderivative of \(0\) with respect to \(y\) is a constant \(C\). Evaluate it at the given limits: $$C(2-1) = C$$ Since the integral evaluated to a constant, the final result is simply: $$\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y = 0$$