Question
(a) Topic-6.9- Integrating Using Substitution
(b) Topic-2.8- The Product Rule
(c) Topic-8.9- Volume with Disc Method: Revolving Around the \(x\)- or \(y\)-Axis
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 \(2\pi\) cubic inches. What is the value of \(c\) for this spinning toy? Ans:
▶️Answer/Explanation
3(a)
\(6x\sqrt{4 – x^2} = 0 \implies x = 0, \, x = 2\)
Area} \(= \int_{0}^{2} 6x\sqrt{4 – x^2} \, dx\)
Let \(u = 4 – x^2\).
\(du = -2x \, dx \implies -\frac{1}{2} \, du = x \, dx\)
\(x = 0 \implies u = 4 – 0^2 = 4\)
\(x = 2 \implies u = 4 – 2^2 = 0\)
\(\int_{0}^{2} 6x\sqrt{4 – x^2} \, dx = \int_{4}^{0} 6\left(-\frac{1}{2}\right)\sqrt{u} \, du= -3\int_{4}^{0} u^{1/2} \, du = 3\int_{0}^{4} u^{1/2} \, du\)\(= 2u^{3/2} \big|_{u=0}^{u=4} = 2 \cdot 8 = 16\)
The area of the region is \(16\) square inches.
3(b)
The cross-sectional circular slice with the largest radius occurs where \(y=c x \sqrt{4 – x^2}\) has its maximum on the interval \(0 < x < 2\).\(\frac{dy}{dx} = \frac{c \, (4 – 2x^2)}{\sqrt{4 – x^2}} = 0 \implies x = \sqrt{2}\)\(x = \sqrt{2} \implies y = c \sqrt{2} \sqrt{4 – (\sqrt{2})^2} = 2c \implies\)\(2c = 1.2 \implies c = 0.6\)
3(c)
\(\text{Volume} = \int_{0}^{2} \pi \big(c x \sqrt{4 – x^2}\big)^2 \, dx
= \pi c^2 \int_{0}^{2} x^2 (4 – x^2) \, dx\)
\(= \pi c^2 \int_{0}^{2} \big(4x^2 – x^4\big) \, dx\)
\(= \pi c^2 \left[ \frac{4}{3}x^3 – \frac{1}{5}x^5 \right]_{0}^{2}\)
\(= \pi c^2 \left(\frac{32}{3} – \frac{32}{5}\right) = \frac{64\pi c^2}{15}\)
\(\frac{64\pi c^2}{15} = 2\pi \implies c^2 = \frac{15}{32} \implies c = \sqrt{\frac{15}{32}}\)