Question
A company designs spinning toys using the family of functions \(y = cx\sqrt{4-x^{2}},\) where c is a positive constant. The figure above shows the region in the first quadrant bounded by the x-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for some c. Each spinning toy is in the shape of the solid generated when such a region is revolved about the x-axis. Both x and y are measured in inches.
(a) Find the area of the region in the first quadrant bounded by the x-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for c = 6.
(b) It is known that, for \(y = cx\sqrt{4-x^{2}},\frac{dy}{dx}=\frac{c\left ( 4-2x^{2} \right )}{\sqrt{4-x^{2}}}.\) For a particular spinning toy, the radius of the largest cross-sectional circular slice is 1.2 inches. What is the value of c for this spinning toy?
(c) For another spinning toy, the volume is 2π cubic inches. What is the value of c for this spinning toy?
Answer/Explanation
Ans:
(a)
\(y = 6\times \sqrt{4-x^{2}}=0\)
x = 0, x = 2
\(A = \int_{0}^{2}6\times \sqrt{4-x^{2}}dx\) u = 4-x2
du = -2xdx
\(A = \int_{4}^{0}-3 \sqrt{u}du=3\int_{0}^{4}u^{\frac{1}{2}}du=3 u^{3/2}\cdot \frac{2}{3}|_{0}^{4}\)
A = 2(43/2 – 03/2) = 2(22.3/2 – 0) = 2(8) = 16
A = 16
(b)
Largest cross section where y is greatest (maximum of y on graph).
Find max:
\(\frac{dy}{dx}=\frac{c(4-2x^{2})}{\sqrt{4-x^{2}}}=0 \rightarrow c(4-2x^{2})=0\)
4 = 2x2
\(x = \sqrt{2}\)
At \(x = \sqrt{2}\), y = 1.2 (largest radius of cross-section equals 1.2, which is max y value)
\(y = cx\sqrt{4-x^{2}}\)
\(1.2 = c\sqrt{2}\left ( \sqrt{4-(\sqrt{2})^{2}} \right )=c\sqrt{2}\left ( \sqrt{4-2} \right )=c\sqrt{2}(\sqrt{2})=2c\)
c = 1.2/2 = 0.6 → c = 0.6
(c)
\(v = \pi \int_{0}^{2}y^{2}dx=\pi \int_{0}^{2}(cx\sqrt{4-x^{2}})^{2}dx = \pi c^{2}\int_{0}^{2}x^{2}(4-x^{2})dx=\pi c^{2}\int_{0}^{2}(4x^{2}-x^{4})dx\)
\(v = \pi c^{2}\left [ \frac{4x^{3}}{3}-\frac{x^{5}}{5}|_{0}^{2} \right ]= \pi c^{2}\left [ \left ( \frac{4(8)}{3}-\frac{32}{5} \right )-(0-0) \right ]\)
\(v = \pi c^{2}\left ( \frac{32(5)}{3(5)}-\frac{32(3)}{5(3)} \right )= \pi c^{2}\left ( \frac{2(32)}{15} \right )== \pi c^{2}\left ( \frac{64}{15} \right )\)
\(2\pi =\pi c^{2}\left ( \frac{64}{15} \right )\)
\(c^{2}=\frac{30}{64}\rightarrow c= \sqrt{\frac{30}{64}}=\frac{\sqrt{30}}{8}\)
Question
Consider the two regions, \(R_{1}\) and \(R_{2}\), shown in the figure below
(A) Is there a value of a > 0 that makes \(R_{1}\) and \(R_{2}\) have equal area? Justify your response.
(B) If the line x = b divides the region \(R_{1}\) into two regions of equal area, express b in terms of a.
(C) Express in terms of a the volume of the solid obtained by revolving the region \(R_{1}\) about the y-axis.
(D) If \(R_{2}\)is the base of a solid whose cross sections perpendicular to the y-axis are squares, find the volume of the solid in terms of a.
Answer/Explanation
(A) Let \(A_{i}\) denote the area of \(R_{i}\).Then,
\(A_{1}=\int_{0}^{a^{2}}\sqrt{x}dx=\left [ \frac{2}{3}x^{3/2} \right ]^{a^{2}}_{0}=\frac{2a^{3}}{3}\)
and
\(A_{2}=\int_{0}^{a}y^{2}dy=\left [ \frac{y^{3}}{3} \right ]^{a}_{0}=\frac{a^{3}}{3}\)
Therefore, you solve \(A_{1}=A_{2}\) for a:
\(\frac{2a^{3}}{3}=\frac{a^{3}}{3}\Leftrightarrow a^{3}=0\Leftrightarrow a=0\)
Thus, there are no solutions a > 0 such that \(A_{1}=A_{2}\)
(B)\(R_{1}\) is divided into two regions of equal area, each expressed by the following integral equations:
\(\frac{A_{1}}{2}=\int_{0}^{b}\sqrt{x}dx=\frac{2}{3}b^{3/2}\)
ans
\(\frac{A_{1}}{2}=\int_{b}^{a^{2}}\sqrt{x}dx=\frac{2}{3}a^{3}-\frac{2}{3}b^{3/2}\)
Upon setting these two expressions equal, you solve for b:
\(\frac{2}{3}b^{3/2}=\frac{2}{3}a^{3}-\frac{2}{3}b^{3/2}\Leftrightarrow 2b^{3/2}=a^{3}\Leftrightarrow b=\frac{a^{2}}{\sqrt[3]{4}}\)
(C)The solid obtained by rotating R1 about the y-axis has cross sections of washers with inner radius \(x=y^{2}\) and outer radius \(x=a^{2}\). Therefore, the volume is expressed as
\(\pi \int_{0}^{a}(a^{4}-y^{4})dy=\pi \left [ a^{4}y-\frac{1}{5}y^{5} \right ]^{a}_{0}=\pi \left ( \frac{5a^{2}}{5}-\frac{a^{5}}{5} \right )=\frac{4\pi a^{5}}{5}\)
(D)At each y belonging to [0, a], the area of a square cross section is given by \(A(y)=(y^{2})^{2}=y^{4}\).Therefore, the volume of the solid is just the value of the integral:
\(\int_{0}^{a}y^{4}dy=\frac{1}{5}a^{5}\)
Question
Consider the region R in the first quadrant under the graph of y = cos x from x = 0 to \(x=\frac{\pi }{2}\).
(A) Find the area of R.
(B) What is the volume of the solid obtained by rotating R about the x-axis?
(C) Suppose R is the surface of a concrete slab. If the depth of the concrete at x, where x is given in feet, is \(d(x)=\sin x+1\) find the volume (in cubic feet) of the concrete slab.
Answer/Explanation
(A) The area bounded by the curves is given by
\(\int_{0}^{\pi /2}\cos xdx=[\sin x]^{\pi /2}_{0}=1\)
(B)If you rotate the region about the x-axis, note that each cross section perpendicular to the x-axis is a disc with radius y = cos x. Therefore, you set up the following integral to compute the volume:
\(\pi \int_{0}^{\pi /2}\cos^{2}xdx=\frac{\pi }{2}\int_{0}^{\pi /2}(1+\cos (2x))dx=\frac{\pi }{2}\left [ x+\frac{\sin (2x)}{2} \right ]^{\pi /2}_{0}=\frac{\pi ^{2}}{4} \)
(C)Cross sections perpendicular to the x-axis are rectangles of depth sinx + 1 and width cos x. Therefore, the volume is given by
\(\int_{0}^{\pi /2}(\sin x+1)\cos xdx\)
This integral is computable by many techniques. You will make the substitution u=sin x+1,so that du = cosxdx, and the integral becomes
\(\int_{1}^{2}udu=\frac{1}{2}[4-1]=\frac{3}{2}\) or \(1.5ft^{3}\)