Problem 33 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Q: Evaluate the triple integral: $$\int_{0}^{2} \int_{0}^{4} \int_{y^{2}}^{4} \sqrt{x} dz dx dy$$. A: The value of the triple integral is $$\frac{256}{3}\sqrt{2}$$.

Step 1: Evaluate the inner integral with respect to z

We first focus on the inner integral: $$\int_{y^{2}}^{4} \sqrt{x} dz$$ Since \(\sqrt{x}\) does not contain \(z\), it is treated as a constant during the integration with respect to \(z\). The integral becomes: $$\sqrt{x} \int_{y^{2}}^{4} dz$$ Now, integrate with respect to \(z\): $$\sqrt{x} [z]_{y^{2}}^{4} = \sqrt{x}(4 - y^2)$$

Step 2: Evaluate the middle integral with respect to x

Now we substitute the result of the inner integral into the middle integral and evaluate with respect to \(x\): $$\int_{0}^{4} \sqrt{x}(4 - y^2) dx$$ To integrate this expression, we need to multiply the \(\sqrt{x}\) with the constant \(4-y^2\) and then integrate with respect to \(x\). The integral becomes: $$ (4-y^2) \int_{0}^{4} \sqrt{x} dx$$ Now, integrate with respect to \(x\): $$ (4-y^2) \frac{2}{3} [x^{3/2}]_{0}^{4} = (4-y^2) \frac{2}{3}(4^{3/2} - 0) = 16\sqrt{2}(4-y^2) \frac{1}{3}$$

Step 3: Evaluate the outer integral with respect to y

Finally, we substitute the result of the middle integral into the outer integral and evaluate with respect to \(y\): $$\int_{0}^{2} 16\sqrt{2}(4-y^2) \frac{1}{3} dy$$ To integrate this expression with respect to \(y\), we first multiply the constant \(16\sqrt{2}\frac{1}{3}\) with \((4-y^2)\) and then integrate with respect to \(y\). The integral becomes: $$ 16\sqrt{2} \frac{1}{3} \int_{0}^{2} (4 - y^2) dy$$ Now, integrate with respect to \(y\): $$16\sqrt{2} \frac{1}{3} [4y - \frac{1}{3}y^3]_{0}^{2} = 16\sqrt{2} \frac{1}{3}(8 - \frac{8}{3}) = 16\sqrt{2} \frac{1}{3} \frac{16}{3}$$ The result of the triple integral is: $$\int_{0}^{2} \int_{0}^{4} \int_{y^{2}}^{4} \sqrt{x} dz dx dy = \frac{256}{3}\sqrt{2}$$