Question
| h (feet) | 0 | 2 | 5 | 10 |
| A(h) (square feet) | 50.3 | 14.4 | 6.5 | 2.9 |
A tank has a height of 10 feet. The area of the horizontal cross section of the tank at height h feet is given by the function A, where A (h) is measured in square feet. The function A is continuous and decreases as h increases. Selected values for A (h) are given in the table above.
(a) Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the volume of the tank. Indicate units of measure.
(b) Does the approximation in part (a) overestimate or underestimate the volume of the tank? Explain your reasoning.
(c) The area, in square feet, of the horizontal cross section at height h feet is modeled by the function f given by \(f(h)=\frac{50.3}{e^{0.2h}+h}\) Based on this model, find the volume of the tank. Indicate units of measure.
(d) Water is pumped into the tank. When the height of the water is 5 feet, the height is increasing at the rate of 0.26 foot per minute. Using the model from part (c), find the rate at which the volume of water is changing with respect to time when the height of the water is 5 feet. Indicate units of measure.
Answer/Explanation
Ans:
(a) \(Volume = \int_{0}^{10}A(h)dh\)
≈ (2 – 0) · A(0) + (5 -2) · A(2) + (10 – 5) · A(5)
= 2 · 50.3 + 3.14.4 + 5 · 6.5
= 176.3 cubic feet
(b) The approximation in part (a) is an overestimate because a left Riemann sum is used and A is decreasing.
(c) \(\int_{0}^{10}f(h)dh=101.325338\)
The volume is 101.325 cubic feet.
(d) Using the model, \(V(h)=\int_{0}^{h}f(x)dx.\)
\(\frac{dV}{dt}|_{h=5}=\left [ \frac{dV}{dh}\cdot \frac{dh}{dt} \right ]_{h=5}\)
\(=\left [f(h)\cdot \frac{dh}{dt} \right ]_{h=5}\)
When h = 5, the volume of water is changing at a rate of 1.694 cubic feet per minute.
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 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
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}
$