Question
Let R and S in the figure above be defined as follows: R is the region in the first and second quadrants bounded by the graphs of 2 \(y = 3 -x^{2} \)and y =2^{x} .S is the shaded region in the first quadrant bounded by the two graphs, the x-axis, and the y-axis.
(a) Find the area of S.
(b) Find the volume of the solid generated when R is rotated about the horizontal line y = -1.
(c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is an isosceles right triangle with one leg across the base of the solid. Write, but do not evaluate, an integral expression that gives the volume of the solid.
▶️Answer/Explanation
\(3-x^2=2^x\) when \(x=-1.63658\) and x=1
let \(a=-1.636558\)
(A)area of S =\(\int_{0}^{1}2^xdx+\int_{1}^{\sqrt{3}}\left ( 3-x^2 \right )dx\)
=2.240
(B)Volume = \(\pi \int_{a}^{1}\left ( (3-x^2+1 \right )^2-\left ( (2^x+1)^2 \right )dx\)
=63.106 or 63.107
(C)Volume =\(\frac{1}{2}\int_{a}^{1}(3-x^2-2^x)dx\)
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 a triangle in the xy-plane with vertices at A = (0, 1), B = (2, 3), and C = (3, 1). Let R denote the region that is bounded by the triangle shown in the figure below.
(A) Find the volume of the solid obtained by rotating R about the x-axis.
(B) Find the volume of the solid obtained by rotating R about the y-axis.
(C) Find the volume of the solid having R as its base while cross sections perpendicular to the x-axis are squares.
▶️Answer/Explanation
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 \(1.5ft^{3}\)
Question
Let f(x) = e2x . Let R be the region in the first quadrant bounded by the graph of f, the coordinate axes, and the vertical line x = k, where k > 0. The region R is shown in the figure above.
(a) Write, but do not evaluate, an expression involving an integral that gives the perimeter of R in terms of k.
(b) The region R is rotated about the x-axis to form a solid. Find the volume, V, of the solid in terms of k.
(c) The volume V, found in part (b), changes as k changes. If \(\frac{dk}{dt}=\frac{1}{3}, determine \frac{dV}{t}when k=\frac{1}{2}.\)
▶️Answer/Explanation
Ans:
(a)
\(\frac{dy}{dx}=2e^{2x}\)
\(p = 1 + k + e^{2x}+ \int_{0}^{k}\sqrt{1+(2e^{2x})^{2}}dx\)
(b)
u = 4x
\(\frac{du}{dx}=4\)
\(v = \int_{0}^{k}\pi r^{2}dx=\int_{0}^{k}\pi (e^{2x})^{2}dx=\pi \int_{0}^{k}e^{4x}dx\)
\(= \pi \int_{0}^{k}\frac{1}{4}e^{4}\cdot du=\frac{\pi }{4}\left [ e^{4x} \right ]_{0}^{k}=\frac{\pi }{4}\left ( e^{4k} -e^{4(0)}\right )\)
\(=\frac{\pi }{4}e^{4k}-1\left ( \frac{\pi }{4} \right )\)
\(v=\frac{\pi }{4}e^{4k}- \frac{\pi }{4}\)
(c)
\(\frac{dV}{dt}=\frac{dV}{dk}\cdot \frac{dk}{dt}=\frac{d}{dk}\left ( \frac{\pi }{4}e^{4k}-\frac{\pi }{4} \right )\cdot \left ( \frac{1}{3} \right )=\frac{\pi }{4}(4)e^{4k}\cdot \left ( \frac{1}{3} \right )\)
\(\frac{dV}{dt}=\frac{\pi }{3}e^{4k}|_{k=\frac{1}{2}}=\frac{\pi }{3}\cdot e^{2}\)