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Chapter 16: Problem 9

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y$$

Short Answer

Expert verified
Question: Evaluate the given triple integral: $$\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y$$ Answer: The value of the given triple integral is $$\frac{28}{3}.$$

Step by step solution

01

Integrate with respect to z

First, we will integrate the given function with respect to \(z\). The integration limits for \(z\) are from 1 to \(e\), and the given function is \(\frac{xy^2}{z}\). To find the integral with respect to \(z\), we can write the function as \(xy^2(\int_{1}^{e} \frac{1}{z} dz)\). Integrating \(\frac{1}{z}\) with respect to \(z\), we get: $$\int_{1}^{e} \frac{1}{z} dz = ln|z| \Big|_{1}^{e} = ln|e| - ln|1| = 1$$ Now, we have: $$\int_{-2}^{2} \int_{1}^{2} xy^{2}dx dy$$
02

Integrate with respect to x

Now, we integrate this function with respect to \(x\). The integration limits for \(x\) are from -2 to 2, and the given function is \(xy^2\). To find the integral with respect to \(x\), we can write the function as \(y^2(\int_{-2}^{2} x dx)\). Integrating \(x\) with respect to \(x\), we get: $$\int_{-2}^{2} x dx = \frac{1}{2}x^2 \Big|_{-2}^{2} = 2 - (-2) = 4$$ Now, we have: $$\int_{1}^{2} 4y^{2}dy$$
03

Integrate with respect to y

Finally, we integrate this function with respect to \(y\). The integration limits for \(y\) are from 1 to 2, and the given function is \(4y^2\). Integrating \(4y^2\) with respect to \(y\), we get: $$\int_{1}^{2} 4y^{2} dy = \frac{4}{3}y^3 \Big|_{1}^{2} = \frac{4}{3}(2^3 - 1^3) = \frac{4}{3}(8-1) = \frac{28}{3}$$ So, the value of the given triple integral is: $$\int_{-2}^{2} \int_{1}^{2} \int_{1}^{e} \frac{x y^{2}}{z} d z d x d y = \frac{28}{3}$$

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