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Chapter 16: Problem 9

Evaluate the following iterated integrals. $$\int_{1}^{3} \int_{0}^{2} x^{2} y d x d y$$

Short Answer

Expert verified
Question: Evaluate the iterated integral $$\int_{1}^{3} \int_{0}^{2} x^{2} y dx dy$$ Answer: The value of the iterated integral is $$\frac{32}{3}$$.

Step by step solution

01

Evaluate the inner integral

To evaluate the iterated integral, we start with the inner integral, which is with respect to x: $$\int_{0}^{2} x^{2} y dx$$ We can see that y is like a constant when we integrate with respect to x. So the integral becomes: $$\int_{0}^{2} x^{2} y dx = y \int_{0}^{2} x^{2} dx$$ Now, we integrate x^2 with respect to x: $$ y (\frac{1}{3}x^3 \Big|_0^2)$$
02

Find the result of evaluating the inner integral

Next, we apply the limits of integration for x: $$ y (\frac{1}{3}(2)^3 - \frac{1}{3}(0)^3) = y (\frac{8}{3})$$
03

Evaluate the outer integral with obtained result

Now that we have the result of the inner integral, we plug it into the outer integral and integrate with respect to y: $$\int_{1}^{3} \frac{8}{3} y dy$$
04

Integrate the expression with respect to y

Now, we integrate the expression with respect to y: $$\int_{1}^{3} \frac{8}{3} y dy = \frac{8}{3} \int_{1}^{3} y dy = \frac{8}{3} (\frac{1}{2} y^{2} \Big|_1^3 )$$
05

Apply the limits of integration for y and find the final answer

Finally, we apply the limits of integration for y: $$\frac{8}{3} (\frac{1}{2}(3)^{2} - \frac{1}{2}(1)^{2}) = \frac{8}{3} (\frac{1}{2} (9 - 1)) = \frac{8}{3} (\frac{1}{2}(8))$$ Simplify the expression: $$\frac{8}{3} (\frac{1}{2}(8)) = \frac{8}{3}(4) = \boxed{32/3}$$ The value of the iterated integral is 32/3.

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