The question is simple enough, but I am unable to get the correct answer even after multiple tries. I integrate the volume element $pi*x^2*dy$ between y equals [0, -3].
Let R be the region between the curve $y = -(x-2)^2+1$ and the x-axis. Find the volume of the region obtained by revolving R about the y-axis.
0
$\begingroup$
definite-integrals
solid-of-revolution
-
0Why are you integrating dy? Use vertical slices and integrate dx. – 2017-01-27
-
0Horizontal slices lets me get the area element as a circle. I am not able to imagine the area element if i take vertical slices. – 2017-01-27
-
0Then you must have the wrong graph. – 2017-01-27
-
0The graph would be a inverted parabola with roots at x = (1, 3). If the revolution is around the Y axis then wouldn't the disc element formed be perpendicular to Y axix with radius element in X – 2017-01-27
-
0Oops -- my error -- I misread it. I though it was around the x-axis. But still, use vertical slices, with cylindrical shells, not disks. – 2017-01-27
-
0Is the answer $8\pi/3$?? – 2017-01-27
-
0The quetstion is a MCQ, and 8pi/3 is not an available option. Could you please share your approach. – 2017-01-27
-
0You should read in your book about the shell method. – 2017-01-27
-
0What does MCQ stand for? – 2017-01-27
-
0Multiple Choice Question – 2017-01-27
1 Answers
0
Using vertical slices, the integral is $$\int_{1}^3 2\pi \,r(x)h(x) dx$$ where $$h(x) = -(x-2)^2 + 1$$ and $$r(x) = x$$
-
0By the way, the integral I posted yields an answer of $\dfrac{16\pi}{3}$. Is that the book answer? – 2017-01-27
-
0Got it. You are correct. Apologies. 16pi/3 is an available answer – 2017-01-27
-
0One follow up. Why didn't you integrate between 0 and 1. Isn't the solid implied by the question created between y axis and the curve given by y. – 2017-01-27
-
0Draw the parabola. Then for a typical x between 1 and 3, draw a vertical slice from the $x$-axis to the curve. The height of the slice is the height of the curve above the $x$-axis (i.e., the $y$-value for the chosen x). The width of the slice is $\Delta x$ (which becomes $dx$ in the integral). If you revolve that slice around the $y$-axis, you get what is called a "cylindrical shell". The volume of that shell is its thickness, $\Delta x$, times its surface area which we take as $2\pi\, r(x)\, h(x)$. – 2017-01-27
-
0The radius of the shell corresponding to the vertical slice at a chosen x value (between 1 and 3) is the horizontal distance from the y-axis to the shell -- which is just x. – 2017-01-27