Exercise 14 solutions | Ludek Cizinsky

🔹 15.1.43

Find the volume of the solid that lies under the plane $4x + 6y - 2z + 15 = 0$ and above the rectangle $$R = {(x, y)

-1 \leq x \leq 2, -1 \leq y \leq 1}$$.

Solution

We can use iterated integral to solve this problem. In general, we can write:

\[\begin{aligned} \int_a^b \int_{c}^d f(x, y) dx dy \end{aligned}\]

where $f(x, y) = 2x + 3y + 7.5$, $(a, b) = (-1, 1)$ and $(c, d) = (-1, 2)$. When assigning $a, b, c, d$, pay attention to the order of integration, i.e., $dydx$ vs $dxdy$. Finally, we can write for the first part:

\[\int_{-1}^1 \int_{-1}^2 f(x, y) dx dy = \int_{-1}^1 [x^2 + 3xy + 7.5x]_{x = -1}^{x = 2} dy = \int_{-1}^1 9y + 25.5 dy\]

And for the second part: $\int_{-1}^1 9y + 25.5 dy = [4.5y^2 + 25.5y]_{y = -1}^{y = 1} = 4.5 + 25.5 - (4.5 - 25.5) = 51$

🔸 15.2.31

Find the volume of the following solid. Under the plane $3x + 2y - z = 0$ and above the region enclosed by the parabolas $y = x^2$ and $x = y^2$.

Solution

Again, this is something that we can solve using the iterated integral. We know that the plane is defined as follows $f(x, y) = 3x + 2y$. The core challenge here is to define the region. We know the region is enclosed by two parabolas $y = x^2$ and $x = y^2$. During exam, I suggest that you draw this on the paper, I will use Geogebra instead, see it here. From the drawing, we can infer the definition of the area as follows:

X-range is $x \in (0, 1)$. This is because the two curves intersect when $x = (x^2)^2 = x^4 \Rightarrow x - x^4 = 0 \Rightarrow x(1 - x^3) = 0 \Rightarrow x = 0 \text{ or } x = 1$
Lower border is then defined as $f_1(x) = x^2$ and upper as $f_2(x) = \sqrt{x}$ where $x \in (0, 1)$. For $f_2$, I transformed this equation $x = y^2$ into $y = \sqrt{x}$. I ommited the negative part, i.e., $y = -\sqrt{x}$ which is irrelevant for our border.

Finally, we can write our iterated integral as follows:

\[\int_{x = 0}^{x = 1} \int_{y = x^2}^{y = \sqrt{x}} = 3x + 2y \ dydx\]

Now, we just do what we did in the previous exercise, therefore we start with the first part:

\[\int_{x = 0}^{x = 1} [3xy + y^2]_{y = x^2}^{y = \sqrt{x}} \ dx = \int_{x = 0}^{x = 1} 3x\sqrt{x} + x - 3x^3 - x^4 dx\]

And now the second part:

\[\int_{x = 0}^{x = 1} 3x\sqrt{x} + x - 3x^3 - x^4 dx = [\frac{6}{5}x^{\frac{5}{2}} + \frac{1}{2}x^2 - \frac{3}{4}x^4 - \frac{1}{5}x^5]_{x = 0}^{x = 1} = F(1) - F(0) = \frac{3}{4}\]

where $F(x) = \frac{6}{5}x^{\frac{5}{2}} + \frac{1}{2}x^2 - \frac{3}{4}x^4 - \frac{1}{5}x^5$.

🔸 8.5.10(a)

A type of light bulb is labeled as having an average life time of 1000 hours. It is reasonable to model the probability of failure of these bulbs by an exponential density function with mean $\mu = 1000$. Use this model to find the probability that a bulb

fails within the first 200 hours
burns for more than 800 hours

Solution

In general, we know that exponential density function is given as follows:

\[f_{\lambda}(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } x \geq 0 \\ 0 &\text{if } x < 0 \end{cases}\]

In addition, we know that $E[X] = \mu = 1000$ where $X$ is random variable representing lifetime of a bulb. Finally, we also know from Rasmus’ note that $\lambda = \frac{1}{\mu}$. Therefore, to answer the first question, we need to solve:

\[\begin{aligned} \int_{0}^{200} f_{\lambda}(x) dx = \int_{0}^{200} \lambda e^{-\lambda x} dx \end{aligned}\]

We start by finding the antiderivative:

\[F(x) = -e^{-\lambda x}\]

And then we computee the integral as:

\[\int_{0}^{200} \lambda e^{-\lambda x} dx = F(200) - F(0) = -e^(-\frac{200}{1000}) - (-1) = 1 - e^(-\frac{200}{1000}) \approx .18\]

To answer the second question, we can compute the same integral but with the exception that we will compute it with the upper boundary equal to 800. Then we just subtract it from 1 and obtain the answer for the second question:

\[1 - \int_{0}^{800} \lambda e^{-\lambda x} dx = 1 - (F(800) - F(0)) \approx .44\]

🔸 15.4.29

The joint density function for a pair of random variables $X$ and $Y$ is

\[f(x, y) = \begin{cases} Cx(1 + y) & \text{if } 0 \leq x \leq 1 \text{, } 0 \leq y \leq 2 \\ 0 &\text{ otherwise} \end{cases}\]

a) Find the value of the constant $C$

b) Find $P(X \leq 1, Y \leq 1)$

c) Find $P(X + Y \leq 1)$

Solution

We start by finding the value of $C$ as follows:

\[\begin{aligned} 1 = \int_0^1 \int_0^2 Cx + Cxy \ dy dx \\ 1 = \int_0^1 [Cxy + \frac{1}{2}Cxy^2]_0^2 \ dx \\ 1 = \int_0^1 2Cx + 2Cx \ dx = \int_0^1 4Cx dx \\ 1 = [2Cx^2]_0^1 = 2C \\ \Rightarrow C = \frac{1}{2} \end{aligned}\]

Note that we equate the above double integral with one since we know that probability always must sum to one 1. Second, we proceed to answer what is $P(X \leq 1, Y \leq 1)$:

\[\begin{aligned} P(X \leq 1, Y \leq 1) = \int_0^1 \int_0^1 \frac{1}{2}x + \frac{1}{2}xy \ dydx = \int_0^1 [\frac{1}{2}xy + \frac{1}{4}xy^2]_0^1 = \int_0^1 \frac{3}{4}x \ dx \end{aligned}\]

And now, the second part of it:

\[\begin{aligned} P(X \leq 1, Y \leq 1) = \int_0^1 \frac{3}{4}x = [\frac{3}{8}x^2]_0^1 = \frac{3}{8} \end{aligned}\]

Finally, we can proceed to answer what is $P(X + Y \leq 1)$. In this case, the challenge is to figure out what are the ranges for which we want to integrate. Or in other words, how to define the area. I always start by plotting given function, you can see it here. This shows that the above inequation is true for $X$ such that $X \in (-\infty, 1)$. Since our density function produces some value apart from zero within the range $([0, 1])$, we can conclude that for $X$, we will integrate from 0 to 1. Now, for $Y$, we can reason similarly and say that the lower bound for $Y$ will be 0 as otherwise our density function outputs zero. The upper bound can then be expressed using $y = -x + 1$ (see the graph and should be clear to you why). Notice that we are not doing anything new in here, in exercise 15.2.31, we needed to do pretty similar process before we even started integrating.

So now we plug everything to our equation and compute $P(X + Y \leq 1)$ as follows:

\[P(X + Y \leq 1) = \int_0^1 \int_{y = 0}^{y = -x + 1} \frac{1}{2}x + \frac{1}{2}xy \ dydx = \int_0^1 [\frac{1}{2}xy + \frac{1}{4}xy^2]_{y = 0}^{y = -x + 1} \ dx = \int_0^1 \frac{1}{4}x^3 - x^2 + \frac{3}{4}x \ dx\]

And now the second part:

\[P(X + Y \leq 1) = \int_0^1 \frac{1}{4}x^3 - x^2 + \frac{3}{4}x \ dx = [\frac{1}{16}x^4 - \frac{1}{3}x^3 + \frac{3}{8}x^2]_0^1 = \frac{5}{48}\]

🔍 About the page

Last updated: 06/12/2022
Unless othewise stated, exercises come from the book: James Stewart, Daniel K. Clegg and Saleem Watson: Calculus: Early Transcendentals, Metric Edition, 9th edition
Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by Rasmus Ejlers Møgelberg. Please note that the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.