Double Integrals: Rectangular & General Regions - Calculus 3

1. Evaluate $\int_0^1 \int_0^{\sqrt{1-y^2}} xy \, dx \, dy$ by first sketching the region and determining if changing the order of integration simplifies the computation.
   Answer: $\frac{1}{8}$ and changing order does not simplify The region is bounded by x = √(1-y²), x = 0, y = 0, y = 1, which is a quarter circle. Computing directly: ∫₀¹ [x²y/2]₀^√(1-y²) dy = ∫₀¹ (1-y²)y/2 dy = ∫₀¹ (y-y³)/2 dy = [y²/4 - y⁴/8]₀¹ = 1/4 - 1/8 = 1/8. Changing order gives ∫₀¹ ∫₀^√(1-x²) xy dy dx, which has the same complexity. Choice A has wrong value. Choices B and D incorrectly claim simplification occurs.
2. The region $D$ is enclosed by $x = 0$, $y = 0$, and $x + 2y = 6$. If $\iint_D f(x,y) \, dA = \int_0^a \int_0^{g(x)} f(x,y) \, dy \, dx$, then $a + g(3) =$:
   Answer: $7.5$ When you encounter a double integral over a region defined by boundary lines, you need to carefully identify the region and express it as limits of integration. This tests your ability to set up iterated integrals by finding the correct bounds. First, let's sketch the region $D$. The boundaries are $x = 0$ (y-axis), $y = 0$ (x-axis), and $x + 2y = 6$. To find where $x + 2y = 6$ intersects the axes: when $x = 0$, we get $y = 3$; when $y = 0$, we get $x = 6$. So $D$ is the triangle with vertices at $(0,0)$, $(0,3)$, and $(6,0)$. For the integral $\int_0^a \int_0^{g(x)} f(x,y) \, dy \, dx$, we integrate with respect to $y$ first (from $y = 0$ to some upper bound), then $x$ (from $x = 0$ to $x = a$). Since $x$ ranges from $0$ to $6$, we have $a = 6$. For a fixed $x$-value, $y$ ranges from $0$ up to the line $x + 2y = 6$. Solving for $y$: $y = \frac{6-x}{2}$. Therefore, $g(x) = \frac{6-x}{2}$. Now $g(3) = \frac{6-3}{2} = \frac{3}{2} = 1.5$, so $a + g(3) = 6 + 1.5 = 7.5$. Choice A ($4.5$) likely comes from calculating only $g(3) + 3$. Choice C ($6$) gives just the value of $a$. Choice D ($9$) might result from incorrectly finding $g(3) = 3$ instead of $1.5$. Study tip: Always sketch the region first, then carefully solve the boundary equation for the variable you're integrating over to find the correct limits.
3. Consider the integral $\int_0^4 \int_{\sqrt{y}}^2 e^{x^3} \, dx \, dy$. This integral is difficult to evaluate as written, but becomes manageable when the order of integration is reversed. What is the value of this integral?
   Answer: $\frac{1}{3}(e^8 - 1)$ When you encounter a double integral that's difficult to evaluate in its given order, reversing the order of integration often makes it manageable. This is especially true when the integrand contains functions like $e^{x^3}$, which has no elementary antiderivative with respect to $x$. To reverse the integration order, you need to sketch the region of integration. The original limits tell us: $0 \leq y \leq 4$ and $\sqrt{y} \leq x \leq 2$. The curve $x = \sqrt{y}$ is equivalent to $y = x^2$. So the region is bounded by $y = x^2$, $x = 2$, and $y = 0$. When integrating with respect to $y$ first, the limits become: $0 \leq x \leq 2$ and $0 \leq y \leq x^2$. The integral becomes: $\int_0^2 \int_0^{x^2} e^{x^3} \, dy \, dx$ Since $e^{x^3}$ doesn't depend on $y$, the inner integral is simply: $\int_0^2 e^{x^3} \cdot x^2 \, dx$ Using the substitution $u = x^3$, so $du = 3x^2 dx$, we get: $\frac{1}{3}\int_0^8 e^u \, du = \frac{1}{3}[e^u]_0^8 = \frac{1}{3}(e^8 - 1)$ This confirms answer B is correct. Answer A has $e^8 - e$ instead of $e^8 - 1$, suggesting an error in evaluating the definite integral bounds. Answer C is missing the factor of $\frac{1}{3}$ from the substitution. Answer D uses $\frac{1}{2}$ instead of $\frac{1}{3}$, indicating a mistake in the substitution setup. Strategy tip: Always sketch the region when changing integration order, and look for substitutions that simplify complex exponential expressions.
4. The region $D$ is defined by $0 \leq x \leq 2$ and $x^2 \leq y \leq 4$. If we reverse the order of integration for $\iint_D f(x,y) \, dA$, what are the new limits of integration?
   Answer: $\int_0^4 \int_{\sqrt{y}}^2 f(x,y) \, dx \, dy$ When x = 0, y ranges from 0 to 4. When x = 2, y ranges from 4 to 4. The region is bounded by x = √y (from y = x²) on the left and x = 2 on the right, with y from 0 to 4. Choice A uses wrong upper x-limit. Choice B incorrectly splits the region. Choice C ignores the curved boundary x = √y.
5. Consider the double integral $\int_0^2 \int_{y/2}^{2-y} (x + y) \, dx \, dy$. If we change the order of integration, which expression correctly represents this integral?
   Answer: $\int_0^1 \int_{2x}^{2-x} (x + y) \, dy \, dx + \int_1^2 \int_0^{2-x} (x + y) \, dy \, dx$ The region is bounded by x = y/2, x = 2-y, and 0 ≤ y ≤ 2. These curves intersect at (1,2), so the region must be split at x = 1. For 0 ≤ x ≤ 1: y ranges from 2x to 2-x. For 1 ≤ x ≤ 2: y ranges from 0 to 2-x (since y/2 = x gives y = 2x, but we need y ≥ 0). Choice A has wrong y-limits. Choice B doesn't split the region. Choice D ignores the y = 2x boundary.
6. Consider the double integral $\iint_R (x^2 + y^2) \, dA$ where $R$ is the region bounded by $y = x^2$, $y = 2x^2$, $x = 1$, and $x = 2$. Which of the following represents the correct setup for this integral?
   Answer: $\int_1^2 \int_{x^2}^{2x^2} (x^2 + y^2) \, dy \, dx$ The region R is bounded vertically between y = x² (lower) and y = 2x² (upper) for x values from 1 to 2. This gives the correct setup as the integral from x = 1 to x = 2, with y ranging from x² to 2x². Choice B has the wrong y-limits (should be 1 to 8, not 1 to 4). Choice C reverses the order of the y-limits. Choice D has incorrect y-limits and reverses the x-limits.

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