Questions: Suppose that f(x, y) = y sqrt(x^3 + 1) on the domain D = (x, y) 0 ≤ y ≤ x ≤ 3. Then the double integral of f(x, y) over D is ∫∫D f(x, y) dx dy= Round your answer to four decimal places.

Solution Steps

The given domain $D = \{(x, y) | 0 \le y \le x \le 3\}$ can be rewritten as $D = \{(x, y) | 0 \le x \le 3, 0 \le y \le x\}$. Therefore, the double integral can be written as: $$ \iint_D f(x,y) \, dA = \int_0^3 \int_0^x y \sqrt{x^3+1} \, dy \, dx $$

Integrating with respect to $y$: $$ \int_0^x y \sqrt{x^3+1} \, dy = \sqrt{x^3+1} \left[ \frac{1}{2}y^2 \right]_0^x = \frac{1}{2} x^2 \sqrt{x^3+1} $$

Substituting the result from Step 2 into the outer integral: $$ \int_0^3 \frac{1}{2} x^2 \sqrt{x^3+1} \, dx $$ Let $u = x^3 + 1$, so $du = 3x^2 dx$, and $x^2 dx = \frac{1}{3} du$. When $x=0$, $u=1$. When $x=3$, $u=28$. The integral becomes: $$ \int_1^{28} \frac{1}{2} \sqrt{u} \frac{1}{3} du = \frac{1}{6} \int_1^{28} u^{1/2} du = \frac{1}{6} \left[ \frac{2}{3} u^{3/2} \right]_1^{28} = \frac{1}{9} \left( 28^{3/2} - 1^{3/2} \right) = \frac{1}{9} (28\sqrt{28} - 1) = \frac{1}{9}(56\sqrt{7} - 1) $$

$$ \frac{1}{9} (56\sqrt{7} - 1) \approx \frac{1}{9} (56(2.64575) - 1) \approx \frac{1}{9}(148.162 - 1) \approx \frac{147.162}{9} \approx 16.35133 $$

Final Answer: