Triple Integral Calculation: A Step-by-Step Solution

Hey guys! Today, we're diving into a fun problem: calculating a triple integral. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step, so you can follow along easily. Our mission is to evaluate the triple integral ∫∫∫E ye^(x+y) dV, where E is defined by the bounds 1 ≤ x ≤ 2, 0 ≤ y ≤ 3, and 1 ≤ z ≤ 3. Let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what we're dealing with.

• Triple Integral: A triple integral is an extension of a single and double integral. Instead of integrating over an interval or a region in a plane, we're integrating over a volume in 3D space.
• E: This represents the region over which we're integrating. In our case, E is a rectangular box defined by the inequalities 1 ≤ x ≤ 2, 0 ≤ y ≤ 3, and 1 ≤ z ≤ 3.
• ye^(x+y): This is the function we're integrating. It depends on x and y, which means our result will depend on these variables within the given bounds.
• dV: This represents the differential volume element. Since we're in Cartesian coordinates, dV = dx dy dz (or any permutation of these).

Keywords: Triple Integral, Region E, Cartesian Coordinates, Differential Volume

Setting Up the Integral

Okay, so now we know what we're trying to compute. Let's set up the triple integral with the correct bounds. Since our region E is a rectangular box, the integral will be relatively straightforward to set up:

∫∫∫E ye^(x+y) dV = ∫12 ∫03 ∫13 ye^(x+y) dz dy dx

Notice that we've set the bounds for z as the innermost integral, y as the middle integral, and x as the outermost integral. This is just a choice, and we could integrate in a different order if we wanted to (Fubini's Theorem says we'll get the same result, as long as the function is continuous, which it is).

Keywords: Integral Setup, Integration Order, Fubini's Theorem, Rectangular Box

Evaluating the Integral

Now comes the fun part – actually evaluating the integral! We'll start by integrating with respect to z:

∫13 ye^(x+y) dz = ye^(x+y) ∫13 dz = ye^(x+y) [z]13 = ye^(x+y) (3 - 1) = 2ye^(x+y)

So, our integral now looks like this:

∫12 ∫03 2ye^(x+y) dy dx

Next, we integrate with respect to y. This part requires a bit more work. We'll use integration by parts. Let's set:

• u = y, dv = e^(x+y) dy
• du = dy, v = e^(x+y)

Then, integration by parts gives us:

∫ ye^(x+y) dy = ye^(x+y) - ∫ e^(x+y) dy = ye^(x+y) - e^(x+y) + C

Now, we evaluate this from 0 to 3:

[ye^(x+y) - e^(x+y)]03 = (3e^(x+3) - e^(x+3)) - (0e^(x+0) - e^(x+0)) = 2e^(x+3) + e^x

Remember, we have that factor of 2 in front of the integral, so we multiply this result by 2:

2 * [2e^(x+3) + e^x] = 4e^(x+3) + 2e^x

So, our integral is now:

∫12 (4e^(x+3) + 2e^x) dx

Finally, we integrate with respect to x:

∫ (4e^(x+3) + 2e^x) dx = 4e^(x+3) + 2e^x + C

Evaluating from 1 to 2:

[4e^(x+3) + 2e^x]12 = (4e^(2+3) + 2e^2) - (4e^(1+3) + 2e^1) = (4e^5 + 2e^2) - (4e^4 + 2e)

We can factor out some terms to make it look nicer:

2[2e^5 + e^2 - 2e^4 - e] = 2[2e^5 - 2e^4 + e^2 - e]

Keywords: Integration by Parts, u-substitution, Bounded Solution, Evaluate

Simplifying the Result

Let's try to massage this expression into one of the answer choices provided. We have:

2[2e^5 - 2e^4 + e^2 - e]

None of the given options directly match, so let's go back and check our work. It is very important not to be afraid to check your work to make sure the solution is valid.

After reviewing the Integration by Parts section:

∫ ye^(x+y) dy = ye^(x+y) - ∫ e^(x+y) dy = ye^(x+y) - e^(x+y) + C

Now, we evaluate this from 0 to 3:

[ye^(x+y) - e^(x+y)]03 = (3e^(x+3) - e^(x+3)) - (0e^(x+0) - e^(x+0)) = 2e^(x+3) + e^x

Looks correct. Now, we integrate with respect to x:

∫12 (4e^(x+3) + 2e^x) dx

Which becomes:

[4e^(x+3) + 2e^x]12 = [4e^(2+3) + 2e^2] - [4e^(1+3) + 2e^1] = 4e^5 + 2e^2 - 4e^4 - 2e

Then we can factor to produce the same answer as before:

2[2e^5 - 2e^4 + e^2 - e]

If we factor out an e from each term in the square brackets, we get:

2e[2e^4 - 2e^3 + e - 1]

Now, let's carefully re-examine the possible solutions:

A. 2/3 * [e ^ 6 + e ^ 3 - 1] B. 2/3 * [e ^ 6 - e ^ 3 - 3] C. 2/3 * [e ^ 6 - e ^ 3 - 1] D. 2 * [e ^ 6 - e ^ 3 - 1] E. 2/3 * [e ^ 6 + e ^ 3 - 3]

Unfortunately, it appears that there was an error with the original problem statement. Now, we must derive the correct answer from the original problem statement.

∫∫∫E ye^(x+y) dV = ∫12 ∫03 ∫13 ye^(x+y) dz dy dx

Taking the constant ye^(x+y) out, we have:

∫12 ∫03 ye^(x+y) ( ∫13 1 dz ) dy dx = ∫12 ∫03 2 ye^(x+y) dy dx = ∫12 2 ( ∫03 ye^(x+y) dy ) dx

Let I = ∫03 ye^(x+y) dy, then we can evaluate this integral using integration by parts, where u = y, du = dy, dv = e^(x+y) dy, v = e^(x+y)

I = [ ye^(x+y) ]03 - ∫03 e^(x+y) dy = 3e^(x+3) - (e^(x+3) - e^x) = 2e^(x+3) + e^x

Then we plug this back in to get:

∫12 2 ( 2e^(x+3) + e^x ) dx = ∫12 4e^(x+3) + 2e^x dx = 4e^(x+3) + 2e^x |12 = (4e^5 + 2e^2) - (4e^4 + 2e)

If we factor out a 2 again, we get 2[2e^5 - 2e^4 + e^2 - e].

It can be confirmed with external software that the answer to this problem is indeed 2[2e^5 - 2e^4 + e^2 - e], and that the provided solutions in the problem are incorrect.

Keywords: Software Verification, Problem Error, External Verification, True Solution

Conclusion

Alright, guys! We've successfully tackled this triple integral. Although there may have been an issue with the possible answers, we have confirmed with external software that our final solution is the true answer. Remember, the key is to break down the problem into manageable steps and carefully apply the integration techniques. Keep practicing, and you'll become a triple integral pro in no time! If you guys have any questions, don't hesitate to ask. Keep calm and integrate on!