Home / AP Calculus AB 8.7 Volumes with Cross Sections: Squares and Rectangles- FRQs

AP Calculus AB 8.7 Volumes with Cross Sections: Squares and Rectangles- FRQs - Exam Style Questions

Calc-Ok Question


The shaded region \(R\) is bounded by the graphs of the functions \(f\) and \(g\), where \(f(x)=x^{2}-2x\) and \(g(x)=x+\sin(\pi x)\), as shown in the figure.
(Note: Your calculator should be in radian mode.)
A. Find the area of \(R\). Show the setup for your calculations.
B. Region \(R\) is the base of a solid. For this solid, at each \(x\) the cross section perpendicular to the \(x\)-axis is a rectangle with height \(x\) and base in region \(R\). Find the volume of the solid. Show the setup for your calculations.
C. Write, but do not evaluate, an integral expression for the volume of the solid generated when the region \(R\) is rotated about the horizontal line \(y=-2\).
D. It can be shown that \(g'(x)=1+\pi\cos(\pi x)\). Find the value of \(x\), for \(0<x<1\), at which the line tangent to the graph of \(f\) is parallel to the line tangent to the graph of \(g\).

Most-appropriate topic codes (CED):

TOPIC 8.4: Finding the Area Between Curves Expressed as Functions of \(x\) — part (A)
TOPIC 8.7: Volumes with Cross Sections: Squares and Rectangles — part (B)
TOPIC 8.12: Volume with Washer Method: Revolving Around Other Axes — part (C)
TOPIC 2.2: Defining the Derivative and Using Derivative Notation (derivative as slope of the tangent line) — part (D)
▶️ Answer/Explanation

(A) Area of \(R\)
Intersections occur at \(x=0\) and \(x=3\) (since \(f(0)=g(0)=0\) and \(f(3)=g(3)=3\)). On \([0,3]\), the top curve is \(g\) and the bottom curve is \(f\).
\(\displaystyle \text{Area}=\int_{0}^{3}\!\big(g(x)-f(x)\big)\,dx=\int_{0}^{3}\!\big(x+\sin(\pi x)-(x^{2}-2x)\big)\,dx\)
\(\displaystyle =\int_{0}^{3}\!\big(-x^{2}+3x+\sin(\pi x)\big)\,dx\)
Antiderivative:
\(\displaystyle \int -x^{2}\,dx=-\frac{x^{3}}{3},\quad \int 3x\,dx=\frac{3x^{2}}{2},\quad \int \sin(\pi x)\,dx=-\frac{\cos(\pi x)}{\pi}.\)
Evaluate from \(0\) to \(3\):
\(\displaystyle \left[-\frac{x^{3}}{3}+\frac{3x^{2}}{2}-\frac{\cos(\pi x)}{\pi}\right]_{0}^{3}=\left(-9+\frac{27}{2}+\frac{1}{\pi}\right)-\left(0+0-\frac{1}{\pi}\right)\)
\(\displaystyle =\frac{9}{2}+\frac{2}{\pi}\approx 4.5+0.636619=\boxed{5.137}\ \text{(to 3 d.p.)}.\)

(B) Volume of solid with rectangular cross sections (height \(x\))
Cross-sectional area at \(x\): \(\displaystyle A(x)=\big(\text{base}\big)\times \big(\text{height}\big)=(g(x)-f(x))\times x.\)
\(\displaystyle A(x)=x\big(-x^{2}+3x+\sin(\pi x)\big)=-x^{3}+3x^{2}+x\sin(\pi x).\)
Volume:
\(\displaystyle V=\int_{0}^{3}\!A(x)\,dx=\int_{0}^{3}\!\big(-x^{3}+3x^{2}+x\sin(\pi x)\big)\,dx.\)
Compute each part:
\(\displaystyle \int_{0}^{3}\!-x^{3}\,dx=\left[-\frac{x^{4}}{4}\right]_{0}^{3}=-\frac{81}{4},\qquad \int_{0}^{3}3x^{2}\,dx=\left[x^{3}\right]_{0}^{3}=27.\)
For \(\displaystyle \int_{0}^{3} x\sin(\pi x)\,dx\), use integration by parts with \(u=x,\; dv=\sin(\pi x)dx\):
\(du=dx,\; v=-\dfrac{\cos(\pi x)}{\pi}\Rightarrow \int x\sin(\pi x)\,dx=-\dfrac{x\cos(\pi x)}{\pi}+\dfrac{\sin(\pi x)}{\pi^{2}}.\)
Evaluate \(0\) to \(3\): \(\displaystyle \left[-\frac{x\cos(\pi x)}{\pi}+\frac{\sin(\pi x)}{\pi^{2}}\right]_{0}^{3}=\frac{3}{\pi}.\)
Sum: \(\displaystyle V=\left(-\frac{81}{4}+27\right)+\frac{3}{\pi}=\frac{27}{4}+\frac{3}{\pi}=\boxed{\frac{12+27\pi}{4\pi}\approx 7.705}.\)

(C) Volume when \(R\) is rotated about \(y=-2\)
Use washers. Outer radius \(R(x)=g(x)-(-2)=g(x)+2\). Inner radius \(r(x)=f(x)-(-2)=f(x)+2\).
\(\displaystyle \text{Volume}=\pi\int_{0}^{3}\!\Big(\big(R(x)\big)^{2}-\big(r(x)\big)^{2}\Big)\,dx=\pi\int_{0}^{3}\!\Big((g(x)+2)^{2}-(f(x)+2)^{2}\Big)\,dx.\) (Do not evaluate.)

(D) Parallel tangents for \(0<x<1\)
Tangent lines are parallel \(\iff\) slopes are equal: \(f'(x)=g'(x)\). Here \(f'(x)=2x-2\) and \(g'(x)=1+\pi\cos(\pi x)\).
Solve \(\displaystyle 2x-2=1+\pi\cos(\pi x)\;\Rightarrow\; 2x-3=\pi\cos(\pi x).\)
Define \(h(x)=2x-3-\pi\cos(\pi x)\). Bracket the root:
\(h(0.60)\approx -1.800+0.971=-0.829<0,\quad h(0.70)\approx -1.600+1.848=0.248>0\Rightarrow\) root in \((0.60,0.70)\).
One Newton step from \(x_{0}=0.68\): \(h'(x)=2+\pi^{2}\sin(\pi x)\) so \(h'(0.68)\approx 2+9.8696\times 0.845\approx 10.342\).
\(h(0.68)\approx 0.057\Rightarrow x_{1}=x_{0}-\dfrac{h(0.68)}{h'(0.68)}\approx 0.68-\dfrac{0.057}{10.342}\approx 0.6745.\)
Refining (calculator, radians) gives \(\boxed{x\approx 0.675819}\) (acceptable to three decimals: \(x\approx 0.676\)).

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