Iterated Integrals - Multivariable Calculus

The iterated integral $\int_0^1 \int_y^{\sqrt{y}} f(x,y) \, dx \, dy$ represents integration over a region $R$ . Which of the following correctly describes this region?
Answer: The region bounded by $y = x^2$ , $x = \sqrt{y}$ , with $0 \leq y \leq 1$ For $0 \leq y \leq 1$ , we need $y \leq x \leq \sqrt{y}$ . This requires $y \leq \sqrt{y}$ , which means $y^2 \leq y$ , or $y(y-1) \leq 0$ . This is satisfied when $0 \leq y \leq 1$ . The region is bounded by $x = y$ (equivalently $y = x$ , but since we're thinking of $x$ as a function of $y$ , this is $x = y$ ) and $x = \sqrt{y}$ (equivalently $y = x^2$ ). Choice A incorrectly includes extra boundaries. Choice B incorrectly states the condition exists only when $y \geq 1$ . Choice D incorrectly concludes the region is empty - the inequality $y \leq \sqrt{y}$ holds for $0 \leq y \leq 1$ .
Evaluate $\iint_D y \, dA$ , where $D$ is the region in the first quadrant bounded by the curves $y = x^3$ and $y = 4x$ .
Answer: $\frac{256}{21}$ When evaluating double integrals over regions bounded by curves, you need to first identify the region of integration, then set up appropriate limits based on whether you integrate with respect to x or y first. To find region D, determine where the curves $y = x^3$ and $y = 4x$ intersect. Setting them equal: $x^3 = 4x$ , so $x^3 - 4x = 0$ , which gives $x(x^2 - 4) = 0$ . This yields $x = 0, 2, -2$ . Since we're in the first quadrant, we use $x = 0$ and $x = 2$ . For $0 \leq x \leq 2$ , the line $y = 4x$ is above the cubic $y = x^3$ , so we integrate y from $x^3$ to $4x$ : $\iint_D y \, dA = \int_0^2 \int_{x^3}^{4x} y \, dy \, dx$ First, integrate with respect to y: $\int_{x^3}^{4x} y \, dy = \left[\frac{y^2}{2}\right]_{x^3}^{4x} = \frac{(4x)^2}{2} - \frac{(x^3)^2}{2} = 8x^2 - \frac{x^6}{2}$ Now integrate with respect to x: $\int_0^2 \left(8x^2 - \frac{x^6}{2}\right) dx = \left[\frac{8x^3}{3} - \frac{x^7}{14}\right]_0^2 = \frac{64}{3} - \frac{128}{14} = \frac{64}{3} - \frac{64}{7} = \frac{448-192}{21} = \frac{256}{21}$ Option A ( $\frac{64}{15}$ ) likely comes from computational errors in the integration. Option B ( $\frac{128}{21}$ ) suggests missing a factor of 2 somewhere. Option C ( $\frac{512}{15}$ ) combines both types of errors. Always sketch the region first to verify your integration bounds, and double-check which curve is the upper boundary throughout your interval.
What is the average value of the function $f(x, y) = 12xy^2$ over the triangular region $T$ with vertices $(0, 0)$ , $(1, 0)$ , and $(1, 2)$ ?
Answer: $\frac{32}{5}$ The average value of a function $f$ over a region $T$ is given by $\frac{1}{\text{Area}(T)} \iint_T f(x, y) \, dA$ . The region $T$ is a right triangle with base 1 and height 2, so its area is $\frac{1}{2}(1)(2) = 1$ . The hypotenuse of the triangle is the line connecting $(0,0)$ and $(1,2)$ , which has the equation $y=2x$ . The region can be described by the inequalities $0 \le x \le 1$ and $0 \le y \le 2x$ . We need to compute the integral $\iint_T 12xy^2 \, dA = \int_0^1 \int_0^{2x} 12xy^2 \, dy \, dx$ . The inner integral is $\int_0^{2x} 12xy^2 \, dy = [4xy^3]_0^{2x} = 4x(2x)^3 = 32x^4$ . The outer integral is $\int_0^1 32x^4 \, dx = [\frac{32}{5}x^5]_0^1 = \frac{32}{5}$ . The average value is $\frac{1}{1} \times \frac{32}{5} = \frac{32}{5}$ .
Rewrite the integral $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f(x, y, z) \, dz \, dy \, dx$ by changing the order of integration to $dx \, dy \, dz$ .
Answer: $\int_0^1 \int_0^{1-z} \int_0^{1-y-z} f(x, y, z) \, dx \, dy \, dz$ The original bounds $0 \le x \le 1$ , $0 \le y \le 1-x$ , and $0 \le z \le 1-x-y$ describe a tetrahedron in the first octant with vertices at (0,0,0), (1,0,0), (0,1,0), and (0,0,1). The top surface is the plane $x+y+z=1$ . To change the order to $dx \, dy \, dz$ , we integrate with respect to $x$ first. For fixed $y$ and $z$ , $x$ varies from the $yz$ -plane ( $x=0$ ) to the plane $x=1-y-z$ . So, the inner integral has bounds $0 \le x \le 1-y-z$ . The middle and outer integrals are over the projection of the tetrahedron onto the $yz$ -plane. This projection is the triangle bounded by $y=0$ , $z=0$ , and $y+z=1$ . This triangular region can be described by the bounds $0 \le z \le 1$ and $0 \le y \le 1-z$ . Combining these, the new integral is $\int_0^1 \int_0^{1-z} \int_0^{1-y-z} f(x, y, z) \, dx \, dy \, dz$ .
Consider the iterated integral $\int_0^2 \int_0^{4-2y} (x+y) \, dx \, dy$ . Which of the following represents the same integral with the order of integration reversed?
Answer: $\int_0^4 \int_0^{\frac{4-x}{2}} (x+y) \, dy \, dx$ The original region is $0 \leq y \leq 2$ and $0 \leq x \leq 4-2y$ . The boundary $x = 4-2y$ can be rewritten as $y = \frac{4-x}{2}$ . To find the range of $x$ : when $y = 0$ , $x = 4-2(0) = 4$ ; when $y = 2$ , $x = 4-2(2) = 0$ . So $x$ ranges from 0 to 4. For each $x$ in $[0,4]$ , $y$ ranges from 0 to $\frac{4-x}{2}$ (since $y \geq 0$ and $y \leq \frac{4-x}{2}$ from the constraint $x \leq 4-2y$ ). Therefore, the reversed integral is $\int_0^4 \int_0^{\frac{4-x}{2}} (x+y) \, dy \, dx$ . Choice A has the wrong upper limit for $y$ . Choice C has the limits reversed (should be from 0 to $\frac{4-x}{2}$ , not from $\frac{4-x}{2}$ to 2). Choice D doesn't actually reverse the order of integration.
Let $R$ be the rectangular region $[0, k] \times [0, 1]$ . For what positive value of $k$ does the integral $\iint_R x \sin(\pi y) \, dA$ equal $\frac{8}{\pi}$ ?
Answer: $2\sqrt{2}$ First, set up the iterated integral: $\int_0^1 \int_0^k x \sin(\pi y) \, dx \, dy$ . Evaluate the inner integral with respect to $x$ : $\int_0^k x \sin(\pi y) \, dx = \sin(\pi y) [\frac{1}{2}x^2]_0^k = \frac{1}{2}k^2 \sin(\pi y)$ . Now, evaluate the outer integral with respect to $y$ : $\int_0^1 \frac{1}{2}k^2 \sin(\pi y) \, dy = \frac{1}{2}k^2 [-\frac{1}{\pi}\cos(\pi y)]_0^1 = \frac{1}{2}k^2 (-\frac{1}{\pi}\cos(\pi) - (-\frac{1}{\pi}\cos(0))) = \frac{k^2}{2\pi}(-(-1) - (-1)) = \frac{k^2}{2\pi}(2) = \frac{k^2}{\pi}$ . We are given that this equals $\frac{8}{\pi}$ . So, $\frac{k^2}{\pi} = \frac{8}{\pi}$ , which implies $k^2=8$ . Since $k$ must be positive, $k = \sqrt{8} = 2\sqrt{2}$ .
Consider the double integral $\iint_R xy \, dA$ where $R$ is the region bounded by $y = 2x$ , $y = 6-x$ , and $x = 0$ . If this integral is set up as $\int_a^b \int_{g_1(x)}^{g_2(x)} xy \, dy \, dx$ , what are the values of $a$ , $b$ , $g_1(x)$ , and $g_2(x)$ ?
Answer: $a = 0$ , $b = 2$ , $g_1(x) = 2x$ , $g_2(x) = 6-x$ When setting up a double integral in the form $\int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx$ , you need to determine the region's boundaries and understand how to integrate with respect to $x$ first, then $y$ . First, identify the region $R$ . The three boundary lines are $y = 2x$ , $y = 6-x$ , and $x = 0$ . To find where $y = 2x$ and $y = 6-x$ intersect, set them equal: $2x = 6-x$ , which gives $3x = 6$ , so $x = 2$ . At $x = 0$ , the lines intersect the y-axis at $(0,0)$ and $(0,6)$ . The region is a triangle with vertices at $(0,0)$ , $(0,6)$ , and $(2,4)$ . For integration in the order $dy \, dx$ , $x$ ranges from $0$ to $2$ (so $a = 0$ , $b = 2$ ). For any fixed $x$ in this interval, $y$ ranges from the lower boundary $y = 2x$ to the upper boundary $y = 6-x$ . Therefore, $g_1(x) = 2x$ and $g_2(x) = 6-x$ . Choice A incorrectly uses $g_1(x) = 0$ and includes a minimum function that's unnecessary. Choice B extends $b$ to $3$ , which goes beyond the intersection point, and uses $g_1(x) = 0$ . Choice C uses $b = 6$ , which is far too large and would include regions outside the bounded area. Study tip: Always sketch the region first and find intersection points. The limits of integration should match exactly where your region begins and ends—no more, no less.
The iterated integral $\int_0^2 \int_{x^2}^{4} f(x, y) \, dy \, dx$ is equivalent to which of the following integrals?
Answer: $\int_0^4 \int_0^{\sqrt{y}} f(x, y) \, dx \, dy$ The original integral is over the region $D$ defined by $x^2 \le y \le 4$ and $0 \le x \le 2$ . This region is bounded by the parabola $y=x^2$ , the horizontal line $y=4$ , and the y-axis $x=0$ . To reverse the order of integration, we need to express the bounds of $x$ in terms of $y$ . The curve $y=x^2$ can be written as $x=\sqrt{y}$ since $x \ge 0$ . For a fixed $y$ between 0 and 4, $x$ ranges from the y-axis ( $x=0$ ) to the parabola ( $x=\sqrt{y}$ ). The range for $y$ over the entire region is from 0 to 4. Therefore, the equivalent integral with the order reversed is $\int_0^4 \int_0^{\sqrt{y}} f(x, y) \, dx \, dy$ .
Consider the iterated integral $\int_0^2 \int_0^{4-y^2} f(x,y) \, dx \, dy$ . If this integral is rewritten with the order of integration reversed, which of the following represents the correct limits of integration?
Answer: $\int_0^4 \int_0^{\sqrt{4-x}} f(x,y) \, dy \, dx$ The original region is bounded by $0 \leq y \leq 2$ and $0 \leq x \leq 4-y^2$ . The curve $x = 4-y^2$ can be rewritten as $y^2 = 4-x$ or $y = \sqrt{4-x}$ (taking the positive root since $y \geq 0$ ). The region extends from $x = 0$ to $x = 4$ (when $y = 0$ ), and for each $x$ , $y$ ranges from $0$ to $\sqrt{4-x}$ . Therefore, the reversed integral is $\int_0^4 \int_0^{\sqrt{4-x}} f(x,y) \, dy \, dx$ . Choice B incorrectly includes negative $y$ values. Choice C uses $\sqrt{x}$ instead of $\sqrt{4-x}$ . Choice D doesn't reverse the order of integration.
A solid region $E$ is defined by the inequalities $x^2 + y^2 \le z \le 4$ . Which of the following iterated integrals represents the volume of $E$ ?
Answer: $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+y^2}^4 1 \, dz \, dy \, dx$ The solid $E$ is bounded below by the paraboloid $z = x^2+y^2$ and above by the plane $z=4$ . The volume of $E$ is given by the triple integral $\iiint_E 1 \, dV$ . The bounds for $z$ are from the lower surface to the upper surface: $x^2+y^2 \le z \le 4$ . To find the bounds for $x$ and $y$ , we project the solid onto the $xy$ -plane. The projection is the region where the bounding surfaces intersect, which is $x^2+y^2 = 4$ , a circle of radius 2. Thus, the domain for $(x,y)$ is the disk $x^2+y^2 \le 4$ . In Cartesian coordinates, this disk is described by $-2 \le x \le 2$ and $-\sqrt{4-x^2} \le y \le \sqrt{4-x^2}$ . Combining these gives the iterated integral for the volume: $\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{x^2+y^2}^4 1 \, dz \, dy \, dx$ . Choice C is a double integral for the volume, but is not the requested triple integral setup.
Let $E$ be the solid region in the first octant bounded by the cylinder $x^2 + y^2 = 4$ , the plane $z = y$ , and the $xy$ -plane. Which of the following iterated integrals represents the volume of $E$ ?
Answer: $\int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^y 1 \, dz \, dy \, dx$ The volume is given by $\iiint_E 1 \, dV$ . The solid $E$ is in the first octant, so $x \ge 0$ , $y \ge 0$ , $z \ge 0$ . The solid is bounded below by the $xy$ -plane ( $z=0$ ) and above by the plane $z=y$ . Thus, the bounds for $z$ are $0 \le z \le y$ . The projection of the solid onto the $xy$ -plane is the portion of the disk $x^2+y^2 \le 4$ that is in the first quadrant. In Cartesian coordinates, this region can be described by $0 \le x \le 2$ and $0 \le y \le \sqrt{4-x^2}$ . Combining these, the volume integral is $\int_0^2 \int_0^{\sqrt{4-x^2}} \int_0^y 1 \, dz \, dy \, dx$ .
Evaluate the iterated integral $\int_0^1 \int_0^x e^{x^2} \, dy \, dx$ by first changing the order of integration.
Answer: $\frac{e-1}{2}$ The region is $0 \leq x \leq 1$ and $0 \leq y \leq x$ . Reversing the order: $0 \leq y \leq 1$ and $y \leq x \leq 1$ . So we get $\int_0^1 \int_y^1 e^{x^2} \, dx \, dy$ . Since $e^{x^2}$ doesn't depend on $y$ , we can integrate with respect to $y$ first in the original integral: $\int_0^1 e^{x^2} \int_0^x 1 \, dy \, dx = \int_0^1 e^{x^2} \cdot x \, dx = \int_0^1 x e^{x^2} \, dx$ . Using substitution $u = x^2$ , $du = 2x \, dx$ : $\frac{1}{2}\int_0^1 e^u \, du = \frac{1}{2}[e^u]_0^1 = \frac{e-1}{2}$ . Choice B omits the factor of $\frac{1}{2}$ . Choice C uses $e^2$ instead of $e$ . Choice D incorrectly computes the definite integral.
Evaluate the integral $\int_0^{\pi/2} \int_0^1 \int_0^x y \cos(z) \, dy \, dz \, dx$ .
Answer: $\frac{\pi^3}{48} \sin(1)$ We evaluate the iterated integral from the inside out. The order of integration is $dy$ , then $dz$ , then $dx$ . First, integrate with respect to $y$ : $\int_0^x y \cos(z) \, dy = \cos(z) [\frac{1}{2}y^2]_0^x = \frac{1}{2}x^2 \cos(z)$ . Next, integrate this result with respect to $z$ : $\int_0^1 \frac{1}{2}x^2 \cos(z) \, dz = \frac{1}{2}x^2 [\sin(z)]_0^1 = \frac{1}{2}x^2 (\sin(1) - \sin(0)) = \frac{1}{2}x^2 \sin(1)$ . Finally, integrate this result with respect to $x$ : $\int_0^{\pi/2} \frac{1}{2}x^2 \sin(1) \, dx = \frac{1}{2}\sin(1) [\frac{1}{3}x^3]_0^{\pi/2} = \frac{1}{2}\sin(1) (\frac{1}{3}(\frac{\pi}{2})^3) = \frac{1}{6}\sin(1) (\frac{\pi^3}{8}) = \frac{\pi^3}{48}\sin(1)$ .
Evaluate the iterated integral $I = \int_0^1 \int_y^1 e^{-x^2} \, dx \, dy$ .
Answer: $\frac{1}{2}(1 - e^{-1})$ The integral cannot be evaluated directly as the antiderivative of $e^{-x^2}$ is not an elementary function. The order of integration must be changed. The region of integration is described by $y \le x \le 1$ and $0 \le y \le 1$ . This is a triangle with vertices $(0,0)$ , $(1,0)$ , and $(1,1)$ . Changing the order of integration, the bounds become $0 \le y \le x$ and $0 \le x \le 1$ . The integral becomes $I = \int_0^1 \int_0^x e^{-x^2} \, dy \, dx$ . The inner integral is $\int_0^x e^{-x^2} \, dy = [y e^{-x^2}]_0^x = x e^{-x^2}$ . The outer integral is then $\int_0^1 x e^{-x^2} \, dx$ . Using the substitution $u = -x^2$ , $du = -2x \, dx$ , we get $-\frac{1}{2} \int_0^{-1} e^u \, du = \frac{1}{2} \int_{-1}^0 e^u \, du = \frac{1}{2} [e^u]_{-1}^0 = \frac{1}{2}(e^0 - e^{-1}) = \frac{1}{2}(1 - e^{-1})$ .
Evaluate $\iint_R (x^3 \cos(y) + 2) \, dA$ where $R$ is the disk $x^2+y^2 \le 4$ .
Answer: $8\pi$ When evaluating double integrals over symmetric regions, you can often use symmetry properties to simplify your work significantly. This integral contains two terms: $x^3 \cos(y)$ and $2$ . Let's handle them separately using the linearity of integration: $\iint_R (x^3 \cos(y) + 2) \, dA = \iint_R x^3 \cos(y) \, dA + \iint_R 2 \, dA$ For the first integral, notice that $x^3$ is an odd function of $x$ , and we're integrating over the disk $x^2 + y^2 \leq 4$ , which is symmetric about the y-axis. When you integrate an odd function over a region symmetric about the axis where the variable changes sign, the result is always zero. Therefore: $\iint_R x^3 \cos(y) \, dA = 0$ The second integral is straightforward: $\iint_R 2 \, dA = 2 \cdot \text{Area}(R) = 2 \cdot \pi \cdot 2^2 = 8\pi$ So the total integral equals $0 + 8\pi = 8\pi$ . Answer B ( $4\pi$ ) would result from incorrectly calculating the area as $\pi \cdot 2 = 2\pi$ instead of $\pi \cdot 2^2$ . Answer C ( $0$ ) occurs if you mistakenly think the entire integrand has odd symmetry. Answer D ( $16\pi$ ) might come from doubling the correct area calculation. Study tip: Always check for symmetry properties before diving into complex coordinate transformations. Recognizing when odd functions integrate to zero over symmetric regions can save you significant computation time on exams.

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