Problem 29 - Calculus Early Transcendentals

Question

When converted to an iterated integral, the following double integrals are easier to evaluate in one order than the other. Find the best order and evaluate the integral. $$\iint_{R} 6 x^{5} e^{x^{3} y} d A ; R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 2\\}$$

Step-by-Step Solution

Answer

Based on the step-by-step solution provided above, the double integral $$\int_{0}^{2} \int_{0}^{2} 6x^5 e^{x^3 y} dx dy$$ does not converge and cannot be evaluated in its current form. This is because, in the process of evaluating the integral, we encountered an integral with an undefined natural logarithm term, \(\ln(0)\), which led to a divergent result.

Step 1: Write the integral in the initial order (x then y)

First, let's write the double integral to be evaluated in the order dx dy, meaning we integrate first with respect to x and then with respect to y: $$\int_{0}^{2} \int_{0}^{2} 6x^5 e^{x^3 y} dx dy$$

Step 2: Write the integral in the alternative order (y then x)

Now let's write the double integral to be evaluated in the order dy dx, meaning we integrate first with respect to y and then with respect to x: $$\int_{0}^{2} \int_{0}^{2} 6x^5 e^{x^3 y} dy dx$$

Step 3: Compare the two orders of integration

When we look at both orders, it is evident that the integrand itself doesn't provide any simplification due to a specific order in this problem. Thus, we can choose either order for evaluation, and neither would be significantly easier to evaluate compared to the other. In this case, let's proceed with the original order of integration (x then y) as it doesn't provide any disadvantage.

Step 4: Evaluate the integral with respect to x

Now we'll evaluate the inner integral for x: $$\int_{0}^{2} \left[ \int_{0}^{2} 6x^5 e^{x^3 y} dx \right] dy$$ Using substitution method, let \(u = x^3 y\), so \(\frac{du}{dx} = 3x^2 y\). Thus, \(dx = \frac{du}{3x^2 y}\). Our new integral with respect to u becomes: $$\int_{0}^{2} \left[ \int_{0}^{8y} 6x^3 e^u \frac{du}{3x^2 y} \right] dy = \int_{0}^{2} \left[ \int_{0}^{8y} 2x e^u du \right] dy$$ Now, we can integrate with respect to u: $$\int_{0}^{2} \left[ 2xe^u \Big|_0^{8y}\right] dy = \int_{0}^{2} 2x (e^{8y} - 1) dy$$ We need to convert back to x variables. Since \(u = x^3y\), we have \(x = \sqrt[3]{\frac{u}{y}}\). Substituting this back into the integral, we get: $$\int_{0}^{2} 2\sqrt[3]{\frac{u}{y}} (e^{8y} - 1) dy$$

Step 5: Evaluate the integral with respect to y

Now, we will evaluate the outer integral for y: $$\int_{0}^{2} 2\sqrt[3]{\frac{u}{y}} (e^{8y} - 1) dy$$ This can be split into two integrals: $$2\sqrt[3]{u} \left( \int_{0}^{2} \frac{e^{8y}}{y} dy - \int_{0}^{2} \frac{1}{y} dy \right)$$ The first part of this integral does not have an elementary antiderivative. However, the second part: $$\int_{0}^{2} \frac{1}{y} dy = \ln(y) \Big|_{0}^{2} = \ln(2) - \ln(0)$$ Since \(\ln(0)\) is undefined, the integral is divergent. Hence, the double integral does not converge, and it cannot be evaluated in its current form.