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Chapter 16: Problem 39

Evaluate the following integrals. $$\int_{0}^{2} \int_{0}^{4} \int_{y^{2}}^{4} \sqrt{x} d z d x d y$$

Short Answer

Expert verified
Question: Evaluate the triple integral $$\int_0^2\int_0^4\int_{y^2}^{4} \sqrt{x}\, dz\, dx\, dy$$ Answer: The value of the triple integral is \(\frac{32}{3}\).

Step by step solution

01

Integrate with respect to z

We have: $$\int_{y^2}^{4} \sqrt{x}\, dz$$ Since there is no z-dependency in integrand, the result will be \(\sqrt{x}\) times the range of z: $$\sqrt{x}(4 - y^2)$$
02

Integrate with respect to x

Now, we will integrate the result obtained in step 1 with respect to x: $$\int_{0}^{4} \sqrt{x}(4 - y^2)\, dx$$ To evaluate this integral, first use the substitution rule: $$u = \sqrt{x} \quad\Rightarrow\quad x = u^2 \quad\Rightarrow\quad dx = 2u\, du$$ Now the integral becomes: $$\int_{0}^{2} 2u^3(4 - y^2)\, du$$ Integrate with respect to u: $$2\left[\frac{1}{4}u^4(4 - y^2)\right]_{0}^{2} = 2\left[4 - y^2\right]$$
03

Integrate with respect to y

Finally, we will integrate with respect to y: $$\int_{0}^{2} 2\left[4 - y^2\right]\, dy$$ Integrate with respect to y: $$\left[8y - \frac{2}{3}y^3\right]_{0}^{2}$$ Substitute the limits and find the value: $$\left[8(2) - \frac{2}{3}(2^3)\right] - \left[0\right] = 16 - \frac{16}{3} = \frac{32}{3}$$ Therefore, the value of the triple integral is \(\frac{32}{3}\).

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