Home / AP Calculus BC: 8.9 Volume with Disc Method: Revolving  Around the x- or y-Axis – Exam Style questions with Answer- FRQ

AP Calculus BC: 8.9 Volume with Disc Method: Revolving  Around the x- or y-Axis – Exam Style questions with Answer- FRQ

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

(A) The line segment $A B$ is defined by the equation $y=x+1$ for $0 \leq x \leq 2$, the line segment $B C$ is given by $y=-2 x+7$ for $2 \leq x \leq 3$, while $A C$ is simply $y=1$.

Note that each cross section perpendicular to the $x$-axis is a washer with inner radius 1 , outer radius $x+1$ for $0 \leq x \leq 2$, and outer radius $7-2 x$ for $2 \leq x \leq 3$. Therefore, the volume of the solid is given by the integral:
$
\begin{aligned}
\pi\left(\int_0^2\left[(x+1)^2-1^2\right] d x+\int_2^3\left[(7-2 x)^2-1^2\right] d x\right) & =\pi\left(\left[\frac{(x+1)^3}{3}-x\right]_0^2+\left[-\frac{(7-2 x)^3}{6}-x\right]_2^3\right) \\
& =\pi\left(\left[(9-2)-\frac{1}{3}\right]+\left[\left(-\frac{1}{6}-3\right)+\left(\frac{27}{6}+2\right)\right]\right) \\
& =\pi\left(\left[\frac{21}{3}-\frac{1}{3}\right]+\left[\left(-\frac{1}{6}-\frac{18}{6}\right)+\left(\frac{27}{6}+\frac{12}{6}\right)\right]\right) \\
& =\pi\left(\frac{20}{3}+\frac{20}{6}\right)=\frac{30 \pi}{3}=10 \pi
\end{aligned}
$

(B) Observe that $A B$ may be written as $x=y-1$ and $B C$ as $x=\frac{7-y}{2}$ for $1 \leq y \leq 3$.
Therefore, revolving $R$ about the $y$-axis is the value of the following integral:
$
\pi \int_1^3\left(\frac{7-y}{2}\right)^2-(y-1)^2 d y=\pi\left(\frac{1}{4} \int_1^3(7-y)^2 d y-\int_1^3(y-1)^2 d y\right)
$
Next, you let $u=7-y$ in the first integral above, noting that under this substitution $-d u=d y$. Therefore, the integral becomes
$
\begin{aligned}
\pi\left(-\frac{1}{4} \int_{y=1}^y u^2 d u-\int_1^3(y-1)^2 d y\right) & =\pi\left(-\frac{1}{4}\left[\frac{1}{3} u^3\right]_{y=1}^{y=3}-\left[\frac{1}{3}(y-1)^3\right]_1^3\right) \\
& =\pi\left(\frac{1}{4}\left[-\frac{1}{3}(7-y)^3\right]_1^3-\left[\frac{1}{3}(y-1)^3\right]_1^3\right) \\
& =\pi\left(-\frac{1}{12}\left[4^3-6^3\right]-\frac{1}{3}\left[2^3\right]\right) \\
\text { (Note: } \left.4^3-6^3=4 \cdot 4^2-4 \cdot 9 \cdot 6 .\right) & =\pi\left(-\frac{1}{12}\left[4\left(4^2-9 \cdot 6\right)\right]-\frac{8}{3}\right) \\
& =\pi\left(-\frac{1}{3}(16-54)-\frac{8}{3}\right) \\
& =\pi\left(\frac{38}{3}-\frac{8}{3}\right)=10 \pi
\end{aligned}
$

(C) The area of each square cross section from $x=0$ to $x=2$ is $A_1(x)=(x+1-1)^2=$ $x^2$, while from $x=2$ to $x=3$, the area is given by $A_2(x)=(-2 x+7-1)^2=(6-2 x)^2=4\left(x^2-\right.$ $6 x+9)$. Therefore, the volume of the described solid is
$
\begin{aligned}
\int_0^2 A_1(x) d x+\int_2^3 A_2(x) d x & =\int_0^2 x^2 d x+4 \int_2\left(x^2-6 x+9\right) d x \\
& =\left[\frac{1}{3} x^3\right]_0^2+4\left[\frac{1}{3} x^3-3 x^2+9 x\right]_2^3 \\
& =\left(\frac{8}{3}\right)+4\left[(9-27+27)-\left(\frac{8}{3}-12+18\right)\right] \\
& =\left(4 \cdot \frac{2}{3}\right)+4\left[9-\frac{26}{3}\right] \\
& =4\left(9-\frac{24}{3}\right)=4[9-8]=4
\end{aligned}
$

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)

\(\frac{dV}{dt}=\frac{dV}{dk}\cdot \frac{dk}{dt}=\frac{d}{dk}\left ( \frac{\pi }{4}e^{4k}-\frac{\pi }{4} \right )\cdot \left ( \frac{1}{3} \right )=\frac{\pi }{4}(4)e^{4k}\cdot \left ( \frac{1}{3} \right )\)

\(\frac{dV}{dt}=\frac{\pi }{3}e^{4k}|_{k=\frac{1}{2}}=\frac{\pi }{3}\cdot e^{2}\)

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   \(2y = 3 -x^{2}\) 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 =\(\frac{1}{2}\int_{a}^{1}(3-x^2-2^x)dx\)

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