Home / AP Calculus AB : 8.8 Volumes with Cross Sections: Triangles and  Semicircles- Exam Style questions with Answer- FRQ

AP Calculus AB : 8.8 Volumes with Cross Sections: Triangles and  Semicircles- Exam Style questions with Answer- FRQ

Question:

Let R be the region enclosed by the graph of f(x) = x4 – 2.3x3 + 4 and the horizontal line y = 4, as shown in the figure above.
(a) Find the volume of the solid generated when R is rotated about the horizontal line y = -2.
(b) 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 a leg in R. Find the volume of the solid.
(c) The vertical line x = k divides R into two regions with equal areas. Write, but do not solve, an equation involving integral expressions whose solution gives the value k.

▶️Answer/Explanation

Ans:

(a)

\(f(x)= 4| _{x = 0 or 2.3} \)               \(\pi \int_{0}^{2.3}\left ( (2+4)^{2}-(2+f(x))^{2} \right )dx = 98.868 units^{3}\)

\(V = \pi \left ( R^{2}-r^{2} \right )\)

(b)

\(A = \frac{1}{2}bh\)

\(A = \frac{1}{2}b^{2}\)

b = 4 – f(x)

\(\frac{1}{2}\int_{0}^{2.3}(4-p(x))^{2}dx= 3.574 units^{3}\)

(c)

\(\int_{0}^{k}(4-f(x))dx= \frac{1}{2}\int_{0}^{2.3}\left ( 4-f(x) \right )dx\)

Question:

The inside of a funnel of height 10 inches has circular cross sections, as shown in the figure above. At height h, the radius of the funnel is given by \(r = \frac{1}{20}\left ( 3+h^{2} \right ),\) where 0 ≤ h ≤  10. The units of r and h are inches.
(a) Find the average value of the radius of the funnel.
(b) Find the volume of the funnel.
(c) The funnel contains liquid that is draining from the bottom. At the instant when the height of the liquid is h = 3 inches, the radius of the surface of the liquid is decreasing at a rate of \(\frac{1}{5}\) inch per second. At this instant, what is the rate of change of the height of the liquid with respect to time? 

▶️Answer/Explanation

Ans:

(a)

\(r_{avg}= \frac{1}{200}\int_{0}^{10}\left ( 3+h^{2} \right )dh\)

\(= \frac{1}{200}\left ( 3h+\frac{h^{3}}{3} \right )_{0}^{10}\textrm{]}\)

\(= \frac{1}{200}\left ( 30+\frac{1000}{3} \right )\)

\(= \frac{109}{60} inches.\)

(b)

\(v = \frac{\pi }{400}\int_{0}^{10}(3+h)^{2}dh\)

\(= \frac{\pi }{400}\int_{0}^{10}(9+6h^{2}+h^{4})dh\)

\(= \frac{\pi }{400}(9h+2h^{3}+\frac{h^{5}}{5})_{0}^{10}\textrm{]}\)

\(= \frac{\pi }{400}(90+2000+20000)\)

\(= \frac{2209\pi }{40}in^{3}\)

(c)

\(r = \frac{1}{20}(3+h^{2})\)

\(\frac{dr}{dt}= \left ( \frac{h}{10} \right )\frac{dh}{dt}\)

\(-\frac{1}{5}= \frac{3}{10}\frac{dh}{dt}\)

\(\frac{dh}{dt}= -\frac{2}{3}\) inch per second

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)

\(v = \pi \int_{0}^{2}y^{2}dx=\pi \int_{0}^{2}(cx\sqrt{4-x^{2}})^{2}dx = \pi c^{2}\int_{0}^{2}x^{2}(4-x^{2})dx=\pi c^{2}\int_{0}^{2}(4x^{2}-x^{4})dx\)

\(v = \pi c^{2}\left [ \frac{4x^{3}}{3}-\frac{x^{5}}{5}|_{0}^{2} \right ]= \pi c^{2}\left [ \left ( \frac{4(8)}{3}-\frac{32}{5} \right )-(0-0) \right ]\)

\(v = \pi c^{2}\left ( \frac{32(5)}{3(5)}-\frac{32(3)}{5(3)} \right )= \pi c^{2}\left ( \frac{2(32)}{15} \right )== \pi c^{2}\left ( \frac{64}{15} \right )\)

\(2\pi =\pi c^{2}\left ( \frac{64}{15} \right )\)

\(c^{2}=\frac{30}{64}\rightarrow c= \sqrt{\frac{30}{64}}=\frac{\sqrt{30}}{8}\)

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