Video: Evaluating a Definite Triple Integral

Evaluate the triple integral ∫_(0) ^(3) ∫_(0) ^(2) ∫_(0) ^(1) 𝑥𝑦𝑧 d𝑥 d𝑦 d𝑧.

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Video Transcript

Evaluate the triple integral.

We use a triple integral to integrate over a three-dimensional region. And whilst it can look pretty intimidating, the process itself is fairly straightforward. We’ll begin by integrating 𝑧 with respect to 𝑧 within the limits of one and zero. We’ll then integrate 𝑦 with respect to 𝑦 between two and zero. And finally, we’ll integrate 𝑥 with respect to 𝑥 between three and zero.

So let’s begin with then integrating 𝑧 with respect to 𝑧. The integral of 𝑧 with respect to 𝑧 is 𝑧 squared over two. And with no limits, we have that constant of integration. Remember, to integrate, we add one to the exponent. And then, we divide by the value of this new exponent. Since we’re integrating between the limits one and zero, we’re going to need to substitute these values into our expression and find the difference. That’s one squared divided by two minus zero squared divided by two, which is one-half. And we’re now left with a double integral.

We’re now going to integrate 𝑦 with respect to 𝑦 between the limits two and zero. The integral of 𝑦 with respect to 𝑦 is 𝑦 squared divided by two. And between our limits, it’s two squared divided by two minus zero squared divided by two, which is two. And we’re left with one simple integral. We have the integral of two times one-half 𝑥 with respect to 𝑥 between three and zero. Now actually, two multiplied by one-half is one. So we’re simply integrating 𝑥 with respect to 𝑥.

The integral of 𝑥 with respect to 𝑥 is 𝑥 squared divided by two. So we need to evaluate these between the limits of three and zero. That’s three squared divided by two minus zero squared divided by two, which is simply 4.5. And we’ve evaluated our triple integral. It’s 4.5.

Now, it’s important to realise that whilst we picked a specific order, that is, we integrated 𝑧 with respect to 𝑧 and then 𝑦 with respect to 𝑦 and 𝑥 with respect to 𝑥, we could have chosen any other order as long as we ensured that each integral was evaluated between the limits given.

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