Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int_{1}^{\ln 8} \int_{1}^{\sqrt{z}} \int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} d x d y d z$$

Step-by-Step Solution

Answer

In this exercise, we have analyzed and provided a step-by-step solution on how to evaluate the given triple integral: $$\int_{1}^{\ln 8} \int_{1}^{\sqrt{z}} \int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} d x d y d z$$ The process involved performing integrations with respect to each variable in order, while using appropriate substitutions and adjustments of the limits. However, due to the complexity of the middle integral, a closed-form solution is not straightforward, requiring the use of numerical or approximation methods to evaluate the remaining parts. Once these methods provide us with values for the integrals with respect to $u$ and $v$, we can then complete the evaluation of the integral with respect to $z$ to find the numerical value of the given triple integral.

Step 1: Evaluate the inner integral w.r.t. \(x\)

First, we will focus on the integral inside: $$\int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} d x$$ The function is separable w.r.t. \(x\), so we can integrate it directly: $$\int_{\ln y}^{\ln 2 y} e^{x+y^{2}-z} dx = [\frac{1}{1}* e^{1*(x+y^{2}-z)}]_{\ln y}^{\ln 2y} = e^{(\ln 2y+y^2-z)} - e^{(\ln y+y^2-z)}$$ Now, the integral has become: $$\int_{1}^{\ln 8} \int_{1}^{\sqrt{z}} (e^{(\ln 2y+y^2-z)} - e^{(\ln y+y^2-z)}) d y d z$$

Step 2: Evaluate the integral w.r.t. \(y\)

Next, we will evaluate the middle integral: $$\int_{1}^{\sqrt{z}} (e^{(\ln 2y+y^2-z)} - e^{(\ln y+y^2-z)}) d y$$ For easier computation, let's split the integral into two and focus on the first part: $$\int_{1}^{\sqrt{z}} e^{(\ln 2y+y^2-z)} dy$$ To evaluate this integral, we perform substitution. Take \(u = \ln 2 y + y^2- z\), so \(du = (\frac{2}{2y} + 2y) dy\). Solving for \(y\) in \(u = \ln 2 y + y^2-z\), we get $$y = \frac{2}{2-\sqrt{4-4\left(-z+u\right)}}$$ Replacing \(y\) with the value above in the expression for \(du\), we find that \(du = \frac{4}{\sqrt{4-4\left(-z+u\right)}} dy.\) From this, we can write \(dy = \frac{\sqrt{4-4\left(-z+u\right)}}{4} du\) Now, the integral becomes: $$ \int e^{u} \frac{\sqrt{4-4\left(-z+u\right)}}{4} du = \frac{1}{4} \int e^{u} \sqrt{4-4\left(-z+u\right)} du$$ The limits of integration need to be updated based on the new variable \(u\). When \(y = 1\), \(u = \ln2 +1^2 -z\). When \(y =\sqrt{z}\), \(u = \ln 2 \sqrt{z}+z -z = \ln 2\sqrt{z}\). Now, substitute back the value of \(u\): $$\frac{1}{4}[\int_{\ln2 +1^2 -z}^{\ln 2\sqrt{z}} e^{u} \sqrt{4-4\left(-z+u\right)} du]$$ Now, let's focus on the integral part: $$\int_{1}^{\sqrt{z}} e^{(\ln y+y^2-z)} dy$$ Similarly, let's take \(v = \ln y + y^2- z\), Thus \(dv = (\frac{1}{y} + 2y) dy\). We can solve for \(y\) in terms of \(v\) in a similar way as before. Also, we need to update the limits of integration accordingly. The final integral would be: $$\int e^{v} \frac{\sqrt{4-4\left(-z+v\right)}}{4} dv$$ $$\frac{1}{4}\int_{1^2 -z}^{z-\sqrt{z}} e^{(v)} \sqrt{4-4\left(-z+v\right)} d v$$ Now we have: $$\int_{1}^{\ln 8} \left[\frac{1}{4}\int_{\ln2 +1^2 -z}^{\ln 2\sqrt{z}} e^{(u)} \sqrt{4-4\left(-z+u\right)} d u - \frac{1}{4}\int_{1^2 -z}^{z-\sqrt{z}} e^{(v)} \sqrt{4-4\left(-z+v\right)} d v \right]d z$$

Step 3: Evaluate the integral w.r.t. \(z\)

At this point, you could use numerical or approximation methods to find the value of the integrals w.r.t. \(u\) and \(v\), given that the integrals appear quite complicated and do not have a simple closed form. Once those values are determined, you can compute the final integral w.r.t. \(z\): $$\int_{1}^{\ln 8} \left[\frac{1}{4}\int_{\ln2 +1^2 -z}^{\ln 2\sqrt{z}} e^{(u)} \sqrt{4-4\left(-z+u\right)} d u - \frac{1}{4}\int_{1^2 -z}^{z-\sqrt{z}} e^{(v)} \sqrt{4-4\left(-z+v\right)} d v \right]d z$$ Perform the final integration, and you would get the value of the given triple integral.