📝 Disk method volume revolving about x-axis (34 MCQs)

📖 Explanation: This is a Easy question. The method of disks states that when a region under $y=f(x)$ is revolved around the x-axis, the cross-sectional area at any point x is a circle with radius $f(x)$, giving area $\pi[f(x)]^2$. The volume is the integral of this area from a to b. Therefore, the correct formula includes the $\pi$ outside the integral and squares the function inside.

📖 Explanation: This tests Easy. For the disk method, the radius of each disk is the function value, $f(x)$. The area of the disk is $\pi [f(x)]^2$. The student's expression $\pi \int_0^4 \sqrt{x} \, dx$ would give the volume of a solid where the radius is $\sqrt{\sqrt{x}}$, which is incorrect. The correct integral is $\pi \int_0^4 x \, dx$. The error is specifically forgetting to square the radius function.

📖 Explanation: This is a Easy question. When revolving a region bounded above by a curve and below by the x-axis about the x-axis, the radius of the disk at any point x is simply the distance from the x-axis to the curve, which is the function value $f(x) = 4 - x^2$. The function is non-negative on the relevant interval, so the radius is $4 - x^2$.

📖 Explanation: This is a direct Medium of the disk method. The volume is $V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$. Evaluating the integral gives $\pi [x^2/2]_0^4 = \pi (16/2 - 0) = 8\pi$. The key is correctly squaring the function and evaluating the definite integral.

📖 Explanation: This tests Easy of the disk method. A vertical slice of the region at a given x, when revolved about the x-axis, sweeps out a thin circular disk. The disk's radius is the distance from the x-axis to the curve, which is $f(x)$, and its thickness is the infinitesimal width $dx$. A cylindrical shell is generated by revolving a vertical strip about the y-axis, not the x-axis.

📖 Explanation: This is an Medium question. The region is bounded by $y=x^2$, the x-axis ($y=0$), and $x=2$. The disk method gives $V = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx = \pi [x^5/5]_0^2 = \pi (32/5) = 32\pi/5$. The limits of integration are from 0 to 2 because the curve intersects the x-axis at x=0 and is bounded by x=2.

📖 Explanation: This is a conceptual question. The disk method is directly applied when the axis of revolution bounds the region. In its simplest form, when revolving about the x-axis, the region is bounded below by the x-axis itself. While the function must be non-negative, it's not strictly required to be in the first quadrant as long as the axis of revolution is a boundary and the function is non-negative on the interval.

📖 Explanation: This is a direct Medium. The disk method for revolving about the x-axis uses the formula $V = \pi \int_a^b [f(x)]^2 dx$. Here, $f(x) = \sin x$, $a=0$, and $b=\pi$. The correct integral is $\pi \int_0^\pi (\sin x)^2 dx = \pi \int_0^\pi \sin^2 x \, dx$. This is a classic Medium of the disk method with a trigonometric function.

📖 Explanation: This tests the Medium of the disk method. The region is under $y=\sqrt{x}$ from x=0 to x=4. The radius of the disk is $f(x) = \sqrt{x}$. The volume is $V = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx$. Option A is correct. Option B incorrectly uses $\sqrt{x}$ as the integrand without squaring it, and Option C uses $x^2$, which would be the volume if the curve was $y=x$.

📖 Explanation: This is an Easy question. The disk method applies when the region is bounded by the axis of revolution. Here, the region is between two curves $y=x$ and $y=2x$, neither of which is the x-axis (the axis of revolution). Revolving this region creates a solid with a hole, requiring the washer method, which subtracts the inner radius from the outer radius. The student's error is using the disk method instead of the washer method.

📖 Explanation: This is an Medium question requiring a trigonometric integral. The volume is $V = \pi \int_0^{\pi/2} \cos^2 x \, dx$. Using the identity $\cos^2 x = (1+\cos 2x)/2$, the integral becomes $\pi \int_0^{\pi/2} \frac{1+\cos 2x}{2} dx = \frac{\pi}{2} [x + \frac{\sin 2x}{2}]_0^{\pi/2} = \frac{\pi}{2} (\frac{\pi}{2} + 0) = \frac{\pi^2}{4}$. This requires knowledge of trigonometric integration.

📖 Explanation: This tests Easy of the Riemann sum. In the disk method, the region is sliced into thin vertical strips of width $\Delta x$. When revolved, each strip becomes a disk of thickness $\Delta x$. In the integral, as $\Delta x$ approaches zero, it becomes the differential $dx$, representing the infinitesimal thickness of the disk.

📖 Explanation: The curve $y = \sqrt{4 - x^2}$ is the upper half of a circle of radius 2. Revolving this region about the x-axis generates a sphere. The volume should be $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2)^3 = \frac{32\pi}{3}$. Using the disk method: $V = \pi \int_{-2}^{2} (\sqrt{4-x^2})^2 dx = \pi \int_{-2}^{2} (4-x^2) dx = \pi [4x - x^3/3]_{-2}^{2} = \pi [(8 - 8/3) - (-8 + 8/3)] = \pi [(16/3) - (-16/3)] = 32\pi/3$.

📖 Explanation: This tests the connection between the function and the disk geometry. The radius of the disk at a given x is $f(x) = \sqrt{x}$. At x=1, the radius is $\sqrt{1} = 1$. The diameter is twice the radius, so $2 \times 1 = 2$. It's a direct Medium, but requires understanding that the diameter is 2r.

📖 Explanation: This is a Easy question. The disk method involves integrating the area of the disks, which is $\pi r^2$, not the circumference $2\pi r$. The circumference is used in the shell method, which is for revolution about the y-axis. All other statements are correct properties of the disk method.

📖 Explanation: This is an Easy question. For $y=x^n$, the volume from 0 to 1 is $V = \pi \int_0^1 (x^n)^2 dx = \pi \int_0^1 x^{2n} dx = \frac{\pi}{2n+1}$. For $n=2$, this is $\frac{\pi}{5}$. For $n=4$, this is $\frac{\pi}{2(4)+1} = \frac{\pi}{9}$. A common mistake is to assume the volume scales with the function, but the squaring inside the integral changes the exponent.

📖 Explanation: This tests the understanding of units in integration. $f(x)$ has units of meters (length). $[f(x)]^2$ has units of square meters (area). The differential $dx$ has units of meters (length). Therefore, the product $[f(x)]^2 dx$ has units of cubic meters. The integral sums these volumes, so the final result is in cubic meters.

📖 Explanation: This is an Medium with an exponential function. The radius is $f(x) = e^x$. The volume is $V = \pi \int_0^2 (e^x)^2 dx = \pi \int_0^2 e^{2x} dx = \pi [\frac{1}{2} e^{2x}]_0^2 = \frac{\pi}{2}(e^4 - e^0) = \frac{\pi}{2}(e^4 - 1)$. This requires careful evaluation of the integral and applying the limits correctly.

📖 Explanation: This is a Hard question involving working backwards. The volume is $V = \pi \int_0^k (\sqrt{x})^2 dx = \pi \int_0^k x dx = \pi [x^2/2]_0^k = \frac{\pi k^2}{2}$. We are given that this equals $8\pi$, so $\frac{\pi k^2}{2} = 8\pi$. Dividing by $\pi$, $\frac{k^2}{2} = 8$, so $k^2 = 16$, and $k = 4$ (since $k>0$).

📖 Explanation: This is an Easy question. The correct integral is $V = \pi \int_0^h (\frac{r}{h}x)^2 dx = \pi \frac{r^2}{h^2} \int_0^h x^2 dx = \pi \frac{r^2}{h^2} [\frac{x^3}{3}]_0^h = \frac{1}{3}\pi r^2 h$. A common mistake is to forget the factor of 1/3 when integrating $x^2$, leading to $\pi \frac{r^2}{h^2} [\frac{x^3}{2}]_0^h = \frac{1}{2}\pi r^2 h$.

📖 Explanation: This tests the connection between the Riemann sum and the disk method. The disk method approximates the solid by a series of thin disks. For the i-th strip of width $\Delta x_i$, the disk has volume $\pi [f(x_i^*)]^2 \Delta x_i$. Summing these gives the Riemann sum, which converges to the integral. The shell method uses $2\pi x f(x) \Delta x$.

📖 Explanation: This is a Easy and Easy question. The disk method requires the radius $f(x)$ to be non-negative on the interval of integration to represent a physical radius. On the interval [-2, 0], the function $y = x^3 - 4x$ is positive (e.g., at x=-1, $y = -1 + 4 = 3$). Wait, let's check: $x^3 - 4x = x(x^2 - 4)$. For x in (-2, 0), x is negative, $x^2 - 4$ is negative, so the product is positive. The function is positive on (-2, 0). The disk method is applicable. The question might be testing if the student checks the sign. If the function were negative, the radius would be $|f(x)|$.

📖 Explanation: This tests the understanding that the cross-section is a disk with radius equal to the function value. The area of a circle is $\pi r^2$. Here the radius is $r = \sqrt{\sin x}$. So the area is $A(x) = \pi (\sqrt{\sin x})^2 = \pi \sin x$. The volume would be the integral of this area from 0 to $\pi$. This question focuses specifically on the area of the cross-section, not the volume.

📖 Explanation: This is a multi-step reasoning question requiring a trigonometric integral. $V = \pi \int_0^\pi \sin^2 x \, dx = \pi \int_0^\pi \frac{1-\cos 2x}{2} dx = \frac{\pi}{2} [x - \frac{\sin 2x}{2}]_0^\pi = \frac{\pi}{2} (\pi - 0) = \frac{\pi^2}{2}$. This requires knowing the identity for $\sin^2 x$ and evaluating the definite integral correctly.

📖 Explanation: This is a conceptual question addressing a common misconception. The disk method is used when the region is bounded by the axis of revolution. The region under the semicircle $y = \sqrt{1-x^2}$ is bounded below by the x-axis, so revolving it creates a solid with no hole (a sphere). The radius of each disk is the y-value of the semicircle. The washer method is used when there's a gap between the axis of revolution and the region.

📖 Explanation: This tests the ability to choose the correct method. The region between $y = \sqrt{x}$ and $y = x^2$ is bounded above and below by curves. The x-axis is not a boundary of the region. Revolving this region about the x-axis creates a solid with a hole, so the washer method is appropriate. The washer method is a generalization of the disk method when the inner radius is not zero. The disk method would only apply if the bottom boundary was the x-axis.

📖 Explanation: This is a conceptual question distinguishing between area and volume. The integral $\int_{-1}^{1} \sqrt{1-x^2} \, dx$ gives the area under the semicircle, which is $\pi/2$. The volume of the sphere generated is $\frac{4}{3}\pi$, which is $\pi$ times the integral of the squared function: $\pi \int_{-1}^{1} (\sqrt{1-x^2})^2 dx = \pi \int_{-1}^{1} (1-x^2) dx = \frac{4\pi}{3}$. This highlights the difference in integrands for area and volume.

📖 Explanation: This tests the interpretation of the geometry. The radius of the base of the cone is the y-value at the end of the interval. The function is $y=2x$. At $x=3$, the y-value is $y = 2(3) = 6$. This is the radius of the largest disk and hence the radius of the base of the cone. The height of the cone is 3.

📖 Explanation: This is a scaling reasoning question. If $f(x)$ is replaced by $2f(x)$, the new volume is $V_{new} = \pi \int_0^2 [2f(x)]^2 dx = \pi \int_0^2 4[f(x)]^2 dx = 4\pi \int_0^2 [f(x)]^2 dx = 4V$. The volume scales with the square of the scaling factor because the radius is squared in the integrand. A common mistake is to think it scales linearly.

📖 Explanation: This is an analysis of method applicability. The region bounded by $y=x^2$ and $y=4$ is between two curves. The x-axis is not a boundary. To use the disk method, the region would need to be bounded by the x-axis. One could use the washer method directly. Alternatively, one could use the disk method for two separate regions if the region is split. For example, the area between $y=x^2$ and the x-axis from -2 to 2, and the area between $y=4$ and the x-axis from -2 to 2. The volume of the solid is the difference of the volumes of these two solids.

📖 Explanation: This tests the ability to interpret an integral. The integrand is $\pi [f(x)]^2$. If the integrand is $\pi x^2$, then $[f(x)]^2 = x^2$, so $f(x) = x$ (assuming positive radius). Therefore, the radius at x=0.5 is indeed 0.5. The student is correct. The question tests if the student can correctly extract the radius function from the integrand.

📖 Explanation: This is a Hard extension question. The disk method requires the radius to be the distance from the axis of revolution to the curve. If the axis is $y=-2$, the distance from the axis to a point on the curve $y=f(x)$ is $f(x) - (-2) = f(x) + 2$. So the radius function becomes $\sqrt{4 - x^2} + 2$. This involves modifying the radius function for a shifted axis and is a more complex Medium of the disk method concept.

📖 Explanation: This tests the understanding of the Riemann sum approximation. For 4 disks from 0 to 1, the width of each disk is $\Delta x = 1/4$. Using right endpoints, the x-values are $x_i = i/4$ for $i=1,2,3,4$. The radius of each disk is $f(x_i) = (x_i)^2 = (i/4)^2$. The volume of each disk is $\pi [f(x_i)]^2 \Delta x = \pi (i/4)^2 (1/4)$. The total approximate volume is the sum of these. Option A is correct. This question connects the integral to its Riemann sum representation.