📝 Volume by slicing method (30 MCQs)

📖 Explanation: This statement reflects a common misconception. The formula $V = \text{base area} \times \text{height}$ only applies to right cylinders (solids with uniform cross-sections). For solids with varying cross-sections, like a pyramid or a cone, this formula fails. The correct approach is to integrate the cross-sectional area function $A(x)$ along the axis, which accounts for changes in the shape and size of the slices. The student's claim is a misinterpretation of the slicing method, which requires an integral, not a simple product.

📖 Explanation: The student's setup is correct. At any $x$ in [0, 2], the vertical distance between the curves $y = x^2$ and $y = 4$ is $4 - x^2$. Since the cross-sections are squares, the side length is this distance, and the area of each square is $(4 - x^2)^2$. The solid is bounded by $x = 0$ and $x = 2$, so the limits of integration are correct. The student has successfully applied the slicing method by integrating the area of the square cross-sections over the interval. This is a direct Medium of the volume formula $V = \int A(x) \, dx$.

📖 Explanation: This problem requires careful setup. The region is bounded by $x = 0$ and $x = y^2$ for $y \in [0, 2]$. The width of the cross-section perpendicular to the y-axis is the horizontal distance between the curves, which is $y^2$. Since the height is twice the width, the height is $2y^2$. The area of the rectangular cross-section is $A(y) = \text{width} \times \text{height} = y^2 \cdot 2y^2 = 2y^4$. The volume is $V = \int_{0}^{2} 2y^4 \, dy = 2 \left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{2 \cdot 32}{5} = \frac{64}{5}$. This is a multi-step problem involving interpreting the geometry and setting up the integral correctly. The distractors come from common errors like not squaring the width or mixing up the axes.

📖 Explanation: The diameter of each semicircle is the vertical distance between the curves, which is $\sin x$. The radius is $\frac{\sin x}{2}$. The area of a semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left( \frac{\sin x}{2} \right)^2 = \frac{\pi}{8} \sin^2 x$. Thus, the volume is $V = \int_{0}^{\pi} \frac{\pi}{8} \sin^2 x \, dx$. This matches option A. The other options arise from incorrectly calculating the area of the semicircle (e.g., forgetting the factor of $1/2$ or using the diameter instead of the radius). This question tests the ability to translate a geometric description into a correct integral expression.

📖 Explanation: The student's setup is correct. The region is bounded above by $y = x$ and below by $y = x^2$ on the interval $[0, 1]$. The vertical distance between the curves is $x - x^2$, which is the side length of the square cross-section. The area of the square is $(x - x^2)^2$. Therefore, the volume is $V = \int_{0}^{1} (x - x^2)^2 \, dx$. The student has correctly applied the slicing method. Option A is incorrect because $x$ is above $x^2$ on this interval. Option B is incorrect because the order of limits does not affect the value of the integral if the sign is handled, but the standard form is from lower to upper. Option D incorrectly relates the area of the square to the side length.

📖 Explanation: This is a classic problem. Place the circle with center at the origin and diameter along the x-axis. The equation of the circle is $x^2 + y^2 = 4$. For a given $x$, the vertical distance across the circle is $2y = 2\sqrt{4 - x^2}$. This is the side length of the square cross-section. The area is $A(x) = (2\sqrt{4 - x^2})^2 = 4(4 - x^2)$. The volume is $V = \int_{-2}^{2} 4(4 - x^2) \, dx = 4 \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = 4 \left( (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) \right) = 4 \left( \frac{16}{3} - (-\frac{16}{3}) \right) = 4 \cdot \frac{32}{3} = \frac{128}{3}$. The correct answer is $\frac{128}{3}$. This problem requires setting up the integral from the geometry of the circle and correctly identifying the side length of the square.

📖 Explanation: This problem requires careful attention to the axis of the cross-sections. Since the cross-sections are perpendicular to the y-axis, we integrate with respect to $y$. The base is bounded by $y = 4 - x^2$ and $y = 0$. Solving for $x$, we get $x = \pm \sqrt{4 - y}$. The length of the cross-section parallel to the x-axis (the side of the square) is the horizontal distance between the two sides of the parabola, which is $2\sqrt{4 - y}$. The area of the square is $(2\sqrt{4 - y})^2$. The volume is $V = \int_{0}^{4} (2\sqrt{4 - y})^2 \, dy$. Option A is incorrect because it uses $4 - y$ as the side length instead of $2\sqrt{4 - y}$. Option C is the setup for cross-sections perpendicular to the x-axis. Option B is the integral for the area of the region, not the volume of the solid.

📖 Explanation: The integral $V = \int_{0}^{2} \pi (4 - y^2) \, dy$ is of the form $\int \pi [r(y)]^2 \, dy$, which represents the volume of a solid of revolution about the x-axis or a solid with circular cross-sections perpendicular to the y-axis. Here, $r(y)^2 = 4 - y^2$, so the radius is $\sqrt{4 - y^2}$. The limits of integration are from $y = 0$ to $y = 2$. This suggests the cross-sections perpendicular to the y-axis are disks whose radii vary with $y$. Option A correctly identifies this. Option B incorrectly states the axis of integration. Option C has incorrect limits. Option D is a specific case that does not match the given integrand, which has a variable radius.

📖 Explanation: The volume is found by integrating the cross-sectional area over the given interval. $V = \int_{1}^{3} (4x + 1) \, dx = \left[ 2x^2 + x \right]_{1}^{3} = (18 + 3) - (2 + 1) = 21 - 3 = 18$. The correct answer is 18. This is a Easy of the slicing formula. Option A (16) comes from incorrectly evaluating the integral. Option C (20) and D (22) are other common arithmetic errors.

📖 Explanation: The volume is $V = \int_{0}^{2} \pi x^2 \, dx = \pi \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8\pi}{3}$. This is a direct Medium of the slicing formula. The distractors come from incorrect integration or arithmetic errors.

📖 Explanation: The base is the circle $x^2 + y^2 = 1$. For a given $x$, the vertical distance across the circle is $2\sqrt{1 - x^2}$. This is the side length of the square. The area is $A(x) = (2\sqrt{1 - x^2})^2 = 4(1 - x^2)$. The volume is $V = \int_{-1}^{1} 4(1 - x^2) \, dx = 4 \left[ x - \frac{x^3}{3} \right]_{-1}^{1} = 4 \left( 1 - \frac{1}{3} - (-1 + \frac{1}{3}) \right) = 4 \left( \frac{2}{3} - (-\frac{2}{3}) \right) = 4 \cdot \frac{4}{3} = \frac{16}{3}$. The correct answer is $\frac{16}{3}$. Option A is correct. This problem is similar to the one with radius 2, but with radius 1.

📖 Explanation: First, find the points of intersection: $x^2 = 4x - x^2$ => $2x^2 - 4x = 0$ => $2x(x - 2) = 0$, so $x = 0$ and $x = 2$. The vertical distance between the curves is $(4x - x^2) - x^2 = 4x - 2x^2$. This is the diameter of the semicircle. The radius is $\frac{4x - 2x^2}{2} = 2x - x^2$. The area of a semicircle is $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2x - x^2)^2 = \frac{\pi}{2} (2x - x^2)^2 = \frac{\pi}{8} (4x - 2x^2)^2$. So the volume is $V = \int_{0}^{2} \frac{\pi}{8} (4x - 2x^2)^2 \, dx$. This matches option A. The other options have incorrect factors of $\pi$ or use the diameter instead of the radius in the area formula.

📖 Explanation: The volume of a solid with known cross-sectional area $A(x)$ is $V = \int_{a}^{b} A(x) \, dx$. Comparing this with the given integral, $A(x) = (2x + 1)^2$. This is a direct conceptual question about the definition of the slicing method.

📖 Explanation: The vertical distance between the curves is $x - x^2$. This is the width of the rectangular cross-section. The height is given as 3. So the area of the cross-section is $A(x) = 3(x - x^2)$. The volume is $V = \int_{0}^{1} 3(x - x^2) \, dx = 3 \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = 3 \left( \frac{1}{2} - \frac{1}{3} \right) = 3 \cdot \frac{1}{6} = \frac{1}{2}$. The correct answer is $\frac{1}{2}$. This problem combines the area between curves with the slicing method for rectangles.

📖 Explanation: The diameter of the semicircle is the vertical distance from the x-axis to the curve, which is $\sqrt{x}$. The radius is $\frac{\sqrt{x}}{2}$. The area of a semicircle is $\frac{1}{2} \pi \left( \frac{\sqrt{x}}{2} \right)^2 = \frac{\pi}{8} x$. So the volume is $V = \int_{0}^{4} \frac{\pi}{8} x \, dx$. This matches option A. The distractors use the wrong expression for the radius or the area of the semicircle.

📖 Explanation: Since cross-sections are perpendicular to the y-axis, we integrate with respect to $y$. The base is bounded by $y = 1 - x^2$ and $y = 0$. Solving for $x$, we get $x = \pm \sqrt{1 - y}$. The side length of the square cross-section is the horizontal distance between these two curves, which is $2\sqrt{1 - y}$. The area is $(2\sqrt{1 - y})^2$. So the volume is $V = \int_{0}^{1} (2\sqrt{1 - y})^2 \, dy$. Option A is the integral for the area of the base, not the volume. Option B is the setup for cross-sections perpendicular to the x-axis. Option D is missing the square in the integrand.

📖 Explanation: The volume is $V = \int_{1}^{3} 2x \, dx = \left[ x^2 \right]_{1}^{3} = 9 - 1 = 8$. This is a straightforward Medium of the slicing method. The distractors are common arithmetic errors.

📖 Explanation: The side length of the square is the vertical distance between the curves, which is $\sin x$. The area of the square is $(\sin x)^2 = \sin^2 x$. The volume is $V = \int_{0}^{\pi} \sin^2 x \, dx$. This is a direct Medium of the slicing method. The distractors use the wrong power of $\sin x$ or confuse the area of a square.

📖 Explanation: This problem is similar to the one with the ellipse. The base is a circle of radius 2, centered at the origin. The diameter is along the x-axis. The equation of the circle is $x^2 + y^2 = 4$. The vertical distance across the circle is $2\sqrt{4 - x^2}$. This is the hypotenuse of the isosceles right triangle. The area of an isosceles right triangle with hypotenuse $h$ is $\frac{h^2}{4}$. So $A(x) = \frac{(2\sqrt{4 - x^2})^2}{4} = \frac{4(4 - x^2)}{4} = 4 - x^2$. The volume is $V = \int_{-2}^{2} (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3}$. The correct answer is $\frac{32}{3}$. This problem tests the ability to relate the geometry of the cross-section to the distance across the base.

📖 Explanation: The volume is $V = \int_{0}^{1} e^x \, dx = \left[ e^x \right]_{0}^{1} = e - 1$. This is a direct Medium of the slicing formula with an exponential area function.

📖 Explanation: The student's setup is incorrect. The region is bounded above by $y = x$ and below by $y = x^2$ on the interval $[0, 1]$. The vertical distance between the curves is $x - x^2$, not $x^2 - x$. The height of the rectangle is 2, so the area is $A(x) = 2(x - x^2)$. The correct integral is $\int_{0}^{1} 2(x - x^2) \, dx$. Option C correctly identifies the error. Option A is incorrect because the sign in the integrand is wrong. Option B is incorrect because the limits are in the correct order. Option D is incorrect because it squares the distance, which would be for square cross-sections.

📖 Explanation: The base is the circle $x^2 + y^2 = 9$. The vertical distance across the circle is $2\sqrt{9 - x^2}$. This is the diameter of the semicircle. The radius of the semicircle is $\sqrt{9 - x^2}$. The area of the semicircle is $\frac{1}{2} \pi (\sqrt{9 - x^2})^2 = \frac{\pi}{2} (9 - x^2)$. The volume is $V = \int_{-3}^{3} \frac{\pi}{2} (9 - x^2) \, dx = \frac{\pi}{2} \left[ 9x - \frac{x^3}{3} \right]_{-3}^{3} = \frac{\pi}{2} \left( (27 - 9) - (-27 + 9) \right) = \frac{\pi}{2} (18 - (-18)) = \frac{\pi}{2} \cdot 36 = 18\pi$. The correct answer is $18\pi$. This problem is a straightforward Medium of the slicing method with a circular base.

📖 Explanation: The base is bounded by $y = 4 - x^2$ and the x-axis. The x-intercepts are $x = \pm 2$. The vertical distance from the x-axis to the parabola is $4 - x^2$. This is the side length of the square. The area is $A(x) = (4 - x^2)^2$. The volume is $V = \int_{-2}^{2} (4 - x^2)^2 \, dx = \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx = \left[ 16x - \frac{8x^3}{3} + \frac{x^5}{5} \right]_{-2}^{2} = (32 - \frac{64}{3} + \frac{32}{5}) - (-32 + \frac{64}{3} - \frac{32}{5}) = 64 - \frac{128}{3} + \frac{64}{5} = \frac{960 - 640 + 192}{15} = \frac{512}{15}$. The correct answer is $\frac{512}{15}$. This problem is a standard example of the slicing method.