Double Integrals: General Regions - Multivariable Calculus

1. Evaluate $\iint_R y \, dA$ where $R$ is the region bounded by the semicircle $y=\sqrt{9-x^2}$ and the x-axis.
   Answer: 18 When you encounter a double integral over a region defined by geometric boundaries, start by visualizing the region and choosing appropriate integration limits. Here, the semicircle $y = \sqrt{9-x^2}$ with the x-axis forms the upper half of a circle with radius 3 centered at the origin. To evaluate $\iint_R y \, dA$, set up the integral with bounds: $x$ ranges from $-3$ to $3$, and for each $x$, $y$ ranges from $0$ to $\sqrt{9-x^2}$. This gives us: $\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} y \, dy \, dx$ First, integrate with respect to $y$: $\int_{0}^{\sqrt{9-x^2}} y \, dy = \left[\frac{y^2}{2}\right]_{0}^{\sqrt{9-x^2}} = \frac{9-x^2}{2}$ Now integrate with respect to $x$: $\int_{-3}^{3} \frac{9-x^2}{2} \, dx = \frac{1}{2}\left[9x - \frac{x^3}{3}\right]_{-3}^{3} = \frac{1}{2}[(27-9) - (-27+9)] = \frac{1}{2}[36] = 18$ Answer choice A ($\frac{9\pi}{2}$) likely comes from confusing this with an area calculation or using polar coordinates incorrectly. Answer choice B (0) would result from incorrectly thinking the integral is zero due to symmetry—but $y$ is always positive in this region. Answer choice D ($\frac{4}{\pi}$) appears unrelated to this geometry. Remember: when integrating $y$ over a region where $y \geq 0$, the result represents the first moment about the x-axis, not just area. Always check that your bounds match the actual geometric region.
2. A region $R$ in the xy-plane is bounded by the lines $y=x+1$, $y=-2x+4$, and the x-axis. Which of the following expressions represents the double integral $\iint_R f(x,y) \, dA$?
   Answer: $\int_{-1}^1 \int_0^{x+1} f(x,y) \, dy \, dx + \int_1^2 \int_0^{-2x+4} f(x,y) \, dy \, dx$ First, sketch the region $R$. The vertices of the triangular region are the intersection points of the lines: $(-1,0)$, $(2,0)$, and $(1,2)$. If we integrate with respect to $y$ first (as a Type I region), we must split the integral into two parts at the x-coordinate of the top vertex, $x=1$. For $-1 \le x \le 1$, the region is bounded below by $y=0$ and above by $y=x+1$. For $1 \le x \le 2$, the region is bounded below by $y=0$ and above by $y=-2x+4$. Therefore, the integral is the sum $\int_{-1}^1 \int_0^{x+1} f(x,y) \, dy \, dx + \int_1^2 \int_0^{-2x+4} f(x,y) \, dy \, dx$. Choice C is also a correct representation of the integral, but as a single integral with the order reversed (a Type II region), not as a sum of integrals in the specified order.
3. Find the volume of the solid in the first octant ($x \ge 0, y \ge 0, z \ge 0$) bounded by the cylinder $y^2 + z^2 = 4$, the plane $x+y=2$, and the coordinate planes.
   Answer: $2\pi - \frac{8}{3}$ The volume is given by the double integral of the height function $z=f(x,y)$ over a region $R$ in the xy-plane. From the cylinder equation, $z = \sqrt{4-y^2}$ (since $z \ge 0$). The region $R$ is bounded by $x+y=2$ and the coordinate planes $x=0, y=0$. This is a triangle with vertices $(0,0), (2,0), (0,2)$. The volume is $V = \iint_R \sqrt{4-y^2} \, dA$. It is easier to integrate with respect to $x$ first. The limits are $0 \le y \le 2$ and $0 \le x \le 2-y$. The integral is $V = \int_0^2 \int_0^{2-y} \sqrt{4-y^2} \, dx \, dy$. The inner integral is $\int_0^{2-y} \sqrt{4-y^2} \, dx = (2-y)\sqrt{4-y^2}$. The outer integral is $\int_0^2 (2-y)\sqrt{4-y^2} \, dy = \int_0^2 2\sqrt{4-y^2} \, dy - \int_0^2 y\sqrt{4-y^2} \, dy$. The first term, $2\int_0^2 \sqrt{4-y^2} \, dy$, is twice the area of a quarter-circle of radius 2, which is $2(\frac{1}{4}\pi(2^2)) = 2\pi$. The second term can be solved with u-substitution ($u=4-y^2$), yielding $\frac{8}{3}$. Thus, the volume is $2\pi - \frac{8}{3}$.
4. The sum of iterated integrals $\int_0^1 \int_0^x f(x,y) \, dy \, dx + \int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx$ is equal to which of the following single iterated integrals?
   Answer: $\int_0^1 \int_y^{2-y} f(x,y) \, dx \, dy$ When you encounter a sum of iterated integrals like this, you're being asked to combine two regions of integration and express them as a single integral with reversed order of integration. The key is to visualize the regions geometrically and find their union. Let's analyze what regions these integrals represent. The first integral $\int_0^1 \int_0^x f(x,y) \, dy \, dx$ describes the region where $0 \leq x \leq 1$ and $0 \leq y \leq x$. The second integral $\int_1^2 \int_0^{2-x} f(x,y) \, dy \, dx$ covers $1 \leq x \leq 2$ and $0 \leq y \leq 2-x$. To reverse the order of integration, we need to express this combined region in terms of $y$ first, then $x$. Looking at both regions together, $y$ ranges from 0 to 1. For a fixed $y$ in $[0,1]$, the $x$-values span from where the line $y = 2-x$ intersects (giving $x = 2-y$) to where $y = x$ intersects (giving $x = y$). Therefore, $x$ ranges from $2-y$ to $y$. This gives us $\int_0^1 \int_{2-y}^y f(x,y) \, dx \, dy$, which is choice B. Choice A has incorrect bounds that don't match our region. Choice C has the $x$-limits reversed, which would give a negative area. Choice D extends $y$ to 2 instead of 1, covering too large a region. Study tip: Always sketch the region of integration first. Converting between orders of integration becomes much clearer when you can visualize the geometric boundaries.
5. Which of the following is equal to the iterated integral $\int_0^2 \int_{y/2}^1 y \sin(x^3 - 1) \, dx \, dy$?
   Answer: $\frac{2}{3}(\cos(1)-1)$ The integral $\int \sin(x^3-1) \, dx$ does not have an elementary antiderivative, so we must reverse the order of integration. The region of integration is given by $0 \le y \le 2$ and $y/2 \le x \le 1$. This is a triangle with vertices at $(0,0)$, $(1,0)$, and $(1,2)$. To reverse the order, we set bounds for $x$ first, then $y$. The region is covered by letting $x$ range from $0$ to $1$. For each $x$, $y$ ranges from the bottom boundary $y=0$ to the top boundary $x=y/2$, which is $y=2x$. The new integral is $\int_0^1 \int_0^{2x} y \sin(x^3 - 1) \, dy \, dx$. Evaluating the inner integral: $\int_0^{2x} y \sin(x^3 - 1) \, dy = \sin(x^3-1)[\frac{y^2}{2}]_0^{2x} = \sin(x^3-1) \frac{4x^2}{2} = 2x^2 \sin(x^3-1)$. The outer integral is $\int_0^1 2x^2 \sin(x^3 - 1) \, dx$. Let $u=x^3-1$, so $du = 3x^2 \, dx$. The integral becomes $\int_{-1}^0 \frac{2}{3} \sin(u) \, du = \frac{2}{3}[-\cos(u)]_{-1}^0 = -\frac{2}{3}(\cos(0) - \cos(-1)) = -\frac{2}{3}(1 - \cos(1)) = \frac{2}{3}(\cos(1) - 1)$.
6. Find the average value of the function $f(x,y) = 6x$ over the region $R$ bounded by the parabola $y=x^2$ and the line $y=2x$.
   Answer: 6 The average value is $\frac{1}{\text{Area}(R)} \iint_R f(x,y) \, dA$. First, find the intersection points: $x^2 = 2x \implies x(x-2)=0$, so $x=0$ and $x=2$. The region is bounded by $0 \le x \le 2$, with $y=2x$ as the upper bound and $y=x^2$ as the lower bound. The area is $\text{Area}(R) = \int_0^2 (2x-x^2) \, dx = [x^2 - \frac{x^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$. Next, calculate the integral of the function: $\iint_R 6x \, dA = \int_0^2 \int_{x^2}^{2x} 6x \, dy \, dx = \int_0^2 6x[y]_{x^2}^{2x} \, dx = \int_0^2 6x(2x-x^2) \, dx = \int_0^2 (12x^2 - 6x^3) \, dx = [4x^3 - \frac{3}{2}x^4]_0^2 = 4(8) - \frac{3}{2}(16) = 32 - 24 = 8$. The average value is $\frac{8}{4/3} = 8 \cdot \frac{3}{4} = 6$.
7. Evaluate $\iint_R (x+y)^2 \, dA$ where $R$ is the region in the xy-plane defined by $|x|+|y| \le 1$.
   Answer: $\frac{2}{3}$ When evaluating double integrals over regions defined by absolute value inequalities like $|x| + |y| \leq 1$, you need to recognize the geometry and exploit symmetry. This region forms a diamond (or square rotated 45°) with vertices at $(1,0)$, $(0,1)$, $(-1,0)$, and $(0,-1)$. Since both the region and integrand $(x+y)^2$ have useful symmetry properties, we can simplify the calculation. The region is symmetric about both axes, and $(x+y)^2$ is an even function with respect to the transformation $(x,y) \mapsto (-x,-y)$. We can evaluate this by breaking the diamond into four triangular regions or by using the substitution $u = x+y$, $v = x-y$. Using the latter approach with Jacobian $\frac{1}{2}$, the region transforms to $|u| + |v| \leq 2$, and our integral becomes: $\frac{1}{2}\iint_{|u|+|v| \leq 2} u^2 \, du \, dv$ By symmetry about the $v$-axis and direct integration, this evaluates to $\frac{2}{3}$. Answer **(B) $\frac{2}{3}$ is correct. Answer (A) $\frac{1}{3}$ likely results from forgetting the Jacobian factor or miscalculating the limits. Answer (C) 1 might come from incorrectly treating this as the area of the region rather than the weighted integral. Answer (D) $\frac{1}{4}$** could result from errors in setting up the integration limits or confusing this with a simpler integral. Strategy tip: When you see absolute value constraints defining regions, immediately sketch the region and look for symmetries that can simplify your calculation before diving into integration.
8. Evaluate the iterated integral $\int_0^8 \int_{\sqrt[3]{y}}^2 e^{x^4} \, dx \, dy$.
   Answer: $\frac{1}{4}(e^{16}-1)$ The inner integral with respect to $x$, $\int e^{x^4} \, dx$, cannot be evaluated in terms of elementary functions. This suggests reversing the order of integration. The region of integration is defined by $0 \le y \le 8$ and $\sqrt[3]{y} \le x \le 2$. This can be rewritten as $y = x^3$, $x=2$, and $y=0$. To reverse the order, we express the limits for $y$ in terms of $x$. The region is bounded by $x$ from $0$ to $2$, and for each $x$, $y$ ranges from $0$ to $x^3$. The new integral is $\int_0^2 \int_0^{x^3} e^{x^4} \, dy \, dx$. The inner integral is $\int_0^{x^3} e^{x^4} \, dy = [y e^{x^4}]_0^{x^3} = x^3 e^{x^4}$. The outer integral is $\int_0^2 x^3 e^{x^4} \, dx$. Using the substitution $u = x^4$, we have $du = 4x^3 \, dx$, so $x^3 \, dx = du/4$. The limits become $u(0)=0$ and $u(2)=16$. The integral becomes $\int_0^{16} e^u \frac{du}{4} = \frac{1}{4}[e^u]_0^{16} = \frac{1}{4}(e^{16}-e^0) = \frac{1}{4}(e^{16}-1)$.
9. Calculate the area of the region in the xy-plane bounded by the graphs of $y=e^x$, $y=\ln x$, $x=1$, and $x=2$.
   Answer: $e^2 - e - 2 \ln 2 + 1$ When finding the area between curves, you need to determine which function is on top throughout the interval and set up the integral as (upper function) - (lower function). First, let's determine the relative positions of $y = e^x$ and $y = \ln x$ on the interval $[1,2]$. At $x = 1$: $e^1 = e \approx 2.718$ and $\ln 1 = 0$. At $x = 2$: $e^2 \approx 7.389$ and $\ln 2 \approx 0.693$. Since $e^x > \ln x$ throughout this interval, we set up: $\text{Area} = \int_1^2 (e^x - \ln x) \, dx = \int_1^2 e^x \, dx - \int_1^2 \ln x \, dx$ The first integral is straightforward: $\int_1^2 e^x \, dx = [e^x]_1^2 = e^2 - e$ For the second integral, use integration by parts with $u = \ln x$ and $dv = dx$: $\int \ln x \, dx = x\ln x - x + C$ So: $\int_1^2 \ln x \, dx = [x\ln x - x]_1^2 = (2\ln 2 - 2) - (1 \cdot 0 - 1) = 2\ln 2 - 2 + 1 = 2\ln 2 - 1$ Therefore: $\text{Area} = (e^2 - e) - (2\ln 2 - 1) = e^2 - e - 2\ln 2 + 1$ This confirms answer C is correct. Answer A incorrectly adds the $\ln x$ integral instead of subtracting it. Answer B has the wrong sign on the constant term. Answer D makes both errors—adding instead of subtracting the $\ln x$ integral and having the wrong constant. Remember: always sketch the region or test key points to determine which curve is on top before setting up your area integral.

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