mark
(10 points)
In this problem, you’re going to find the volume of your own, personal tent. To find the dimension of your tent, choose your name from the following list:
An archived instance of a Calc II forum
Volume of your tent
chowell1
The dimensions of my tent are a = 6 and b = 4. The volume of the base is
6* 4 = 24
The radius of the half circle created by the shock cord is
\sqrt{\frac{6^2}{2^2}+\frac{4^2}{2^2}} = \sqrt{\frac{36}{4}+\frac{16}{4}}
=\sqrt{9+6}=\sqrt{13}
the formula of the circle is r^2 + z^2=13 which can be translated to a formula in terms of z:
r=\sqrt{13-z^2}
Thus, our linear scaling factor s, used to change the size of the base in terms of a ‘slice’, can be calculated at height z:
s = \frac{\sqrt{13-z^2}}{\sqrt{13}}
In order to figure out the cross-sectional area, we multiply the area of the base (4 \times 6 = 24) by the square of the linear scaling factor, or s^2. If we write this out, we can see that the cross-sectional area of the slice at height z is
A(z) = 24 \times \frac{13-z^2}{13}
With this information, we can solve the integral to get the area of the tent:
24\int_0^{\sqrt{13}}\frac{13-z^2}{13} = z - \frac{z^3}{39}|_0^{\sqrt{13}} = 24\times(\sqrt{13}-\frac{(\sqrt{13})^3}{39})
afernan2
with tent dimensions of 5 and 2, we get a base volume of 10 (5*2)
and because of the arch of the tent, we know that from the line from a corner of the base to it’s center is equal to my tent’s height. \frac{2}{2}^2+1^2=c^2
c=2.69...\sqrt{7.25-z^2}=x
using this, we apply our formula and evaluate.
2.69\sqrt{3}/2*(7.25*z^2)|^{2.69}_0=16.88-0=16.88
agooch
The dimensions of my tent are a=5 and b=3. The volume of the base is
5*3 =15
The radius of half of the circle created by the shock cord is
\sqrt{\frac{5}{2}^2+\frac{3}{2}^2} = \sqrt{ \frac{25}{4} - \frac{9}{4}} = \sqrt{ \frac{34}{4}}
The formula of the circle in terms of z is:
r= \sqrt{\frac{25}{4} -z^2 }
With this, we can solve for the area of the tent to be
15\int_{0}^{\frac{5}{2}}{ \frac{25}{4}-z^2 dz}= 15* \frac{125}{12}
audrey
The base of my tent is 5 feet long my 4 feet wide; it looks like so:
From the top, it looks like a rectangle:
Note that the distance from the center of the rectangular base to a vertex is
\sqrt{\left (\frac{5}{2}\right)^2+\left (\frac{4}{2}\right)^2} = \sqrt{\frac{41}{4}}.
Now, if we look at this thing from the side (perpendicular to the plane containing one of the semi-circular pieces of shock cord, we should see something like so:
Note that, at height z, the rectangle is smaller so that the length of the segment from the center to a vertex is just
r = \sqrt{41/4 - z^2}.
Thus, the linear scaling factor s to transform the larger base to the smaller cross-section at height z is
s = \frac{\sqrt{41/4 - z^2}}{\sqrt{41/4}}
In order to figure out the the cross-sectional area, we should multiply the area of the base (5\times4=20) by the square of the linear scaling factor - i.e. s^2. Taking all this into account, the cross-sectional area of the slice at height z is
A(z) = 20 \times \frac{41/4 - z^2}{41/4}.
Thus, the volume is
20 \int_0^{\sqrt{41/3}} \frac{41/4 - z^2}{41/4} \, dz = \frac{20 \sqrt{41}}{3}.
mearing
The dimensions of my tent are a=8 and b=4
The area of the base of my tent is 8*4=32
The radius of the half-circle is:
\sqrt{\frac{8}{2}^2+\frac{4}{2}^2}=\sqrt{20}
The formula for my half-circle is:
r=\sqrt{20-z^2}
My scaling factor is:
s=\frac{\sqrt{20-z^2}}{\sqrt{20}}
My area formula is:
A(z)=32\frac{\sqrt{20-z^2}}{\sqrt{20}}
Volume comes out to:
32\int_0^{\sqrt{20}}\frac{20-z^2}{20}dz=\frac{128\sqrt{5}}{3}
myost
the dimensions of my tent are a=7 and b=6
the area of the base of my tent is 7*6=42
the radius of the half circle is:
\sqrt{\left (\frac{7}{2}\right)^2+\left (\frac{6}{2}\right)^2} = \sqrt{\frac{85}{4}}.
formula for half circle:
r=\sqrt{85/4-z^2}
the scaling factor is:
s = \frac{\sqrt{85/4 - z^2}}{\sqrt{85/4}}
cross sectional area (area of base \times square of linear scaling factor)
A(z) = 42 \times \frac{85/4 - z^2}{85/4}.
Thus, the volume is:
42 \int_0^{\sqrt{85/4}} \frac{85/4 - z^2}{85/4} \, dz = 14 \sqrt{85}
jlajcin
The dimensions of my tent are a=7 and b=3, this area of the base of my tent is 7*3=21
The radius of the half-circle is:
\sqrt{\frac{7}{2}^2+\frac{3}{2}^2}=\sqrt{14.5}
The formula for the half-circle is:
r=\sqrt{14.5-z^2}
The scaling factor is:
s=\frac{\sqrt{14.5-z^2}}{\sqrt{14.5}}
The area formula is:
A(z)=21\frac{\sqrt{14.5-z^2}}{\sqrt{14.5}}
Which makes the volume come out to:
21\int_0^{\sqrt{14.5}}\frac{14.5-z^2}{14.5}dz=53.31
ccase2
The dimensions of my tent are a=6 and b=2
The area of the base of my tent is 6*2=12
The radius of the half-circle is:
\sqrt{\left (\frac{6}{2}\right)^2+\left (\frac{2}{2}\right)^2} = \sqrt{10}.
The formula for the half-circle:
r=\sqrt{10-z^2}
The scaling factor is:
s=\frac{\sqrt{10-z^2}}{\sqrt{10}}
The area formula is:
A(z)=12*\frac{10-z^2}{10}
Thus, the volume is:
12 \int_0^{\sqrt{10}} \frac{10 - z^2}{10} \, dz = 12*\frac{10^{3/2}}{15} \approx 25.32
rtaylor4
The dimensions of my tent are a=8 and b=6
The area of the base of my tent is 8*6=48
The radius of the half-circle is:
\sqrt(((8/2)^2)+((6/2)^2))) = \sqrt(25)
The formula for the half circle is:
r = \sqrt(25-z^2)
My scaling factor is:
s = \sqrt(25-z^2)/sqrt(25)
The area formula is:
A(z) = 48((25-z^2)/25)
The volume is then calculated as:
48\int_0^{\sqrt{25}}\frac{25-z^2}
{25}dz=320
knguyen3
If we slice horizontally, the area of the slice is 7s*4s=28s^2ft^2, as s is the scaling factor as the slice goes from the bottom to the top.
If we slice the tent diagonally, we see that r^2+z^2=\frac{65}{4}, as r is the half-length of the diagonal of a rectangle as sliced horizontally and z is the height from the base to the sliced surface. So,r = \sqrt{\frac{65}{4}-z^2}.
The scaling factor of the sliced areas is the scaling factor of the diagonal length and equals to s=\frac{r}{\frac{\sqrt{65}}{2}}=\frac{\sqrt{\frac{65}{4}-z^2}}{\frac{\sqrt{65}}{2}}=\sqrt{\frac{\frac{65}{4}-z^2}{\frac{65}{4}}}=\sqrt{1-\frac{4}{65}z^2}
The volume of the object is V=28\int_0^\frac{\sqrt{65}}{2}(1-\frac{4}{65}z^2)dz=28(z-\frac{4}{195}z^3)|_0^\frac{\sqrt{65}}{2}=28(\frac{\sqrt{65}}{2}-\frac{4}{195}(\frac{\sqrt{65}}{2})^3)=\frac{28\sqrt{65}}{3}ft^3
knguyen3
I have a question. Why scaling factor is squared?
mark
What’s the area of a square if you scale it by the factor 1/2 in all directions?
knguyen3
Oh, it makes sense. Thank you, Dr.McClure!
tbrincke
The dimensions of my tent are a = 12 and b = 12.
The area of the base is 144
The radius of the semicircle is
\sqrt{\frac{12^2}{2^2}+\frac{12^2}{2^2}} = \sqrt{\frac{144}{4}+\frac{144}{4}} = \sqrt{72}
The formula for my semicircle is
r=\sqrt{72-z^2}
The scaling factor is
s=\frac{\sqrt{72-z^2}}{\sqrt{72}}
The area formula is
A(z)=144*\frac{72-z^2}{72}
The volume comes out to
144\int_0^\sqrt{72}\frac{72-z^2}{72}dz= 144(\sqrt{72}-\frac{72^{\frac{3}{2}}}{216}) = 814.59
gbelk
The dimensions of my tent are a =6 and b= 4. the volume of the base is
6*4=24
the radius of the half circle created by the shock cord is:
\sqrt{(6^2/2^2) +(1^2/2^2)} =\sqrt{(36/4+1/4)}=\sqrt{(9+1/4)}
the formula of the circle is r^2+z^2=9.25 which can be translated to a formula in terms of z:
r=\sqrt{(9.25-z^2)}
Thus, our linear scaling factor s, used to change the size of the base in terms of a ‘slice’, can be calculated at height z:
s=(\sqrt{((37/4)-z^2)})/\sqrt{(37/4)}
In order to figure out the cross-sectional area, we multiply the area of the base (6 * 1=24) by the square of the linear scaling factor, or s^2. if we write this out, we can see the cross-sectional area of the slice at height z is
A(z) = 6 *((37/4)-z^2)/(37/4)
with this information we can sole the integral to get the area of the tent:
6 \int_0^\sqrt{37/4}((37/4)-z^2)/(37/4)=6(z-(4z^3)/111)|_0^{\sqrt{37/4}}= 6\sqrt{37/4}-(24(\sqrt{37/4})^3)/111)
rstahles
The dimensions of my tent are a=10 and b=10. The volume of the base is:
10\times10=100
The radius of the half-circle created by the shock cord is:
\sqrt{\frac{10^2}{2^2}+\frac{10^2}{2^2}} = \sqrt{\frac{100}{4}+\frac{100}{4}}
=\sqrt{25+25}=\sqrt{50}
The formula of the circle is r^2+z^2=50 which can be translated to a formula in terms of z:
r=\sqrt{50-z^2}
Thus, our linear scaling factor s, used to change the size of the base in terms of a ‘slice’, can be calculated at height z:
s = \frac{\sqrt{50-z^2}}{\sqrt{50}}
In order to figure out the cross-sectional area, we multiply the area of the base (10\times10=100) by the square of the linear scaling factor, or s^2. If we write this out, we can see that the cross-sectional area of the slice at height, z, is:
A(z) = 100 \times \frac{50-z^2}{50}
With this information, we can solve the integral to get the area of the tent:
100\int_0^{\sqrt{50}}\frac{50-z^2}{50}dz = z - \frac{z^3}{150}|_0^{\sqrt{50}} = 100\times(\sqrt{50}-\frac{(\sqrt{50})^3}{150})\approx471.40
nhaley
The dimensions of my tent are a = 8 and b = 5.
The area of the base of my tent is 8 * 5 = 40.
The radius of the half circle is:
\sqrt{\frac{8}{2}^2+\frac{5}{2}^2} = \frac{\sqrt{89}}{2}
The formula for my half circle is:
r = \sqrt{\frac{89}{2}-z^2}
My scaling factor is:
s = \frac{\sqrt{\frac{89}{2}-z^2}}{\sqrt{\frac{89}{2}}}
My area formula is:
A(z) = 40\frac{\frac{89}{2}-z^2}{\frac{89}{2}}
Volume comes out to:
40\int_{0}^{\sqrt{\frac{89}{2}}}\frac{\frac{89}{2}-z^2}{\frac{89}{2}}dz = \frac{\sqrt{178}}{3}\approx4.448
fcarrill
The dimensions of my tent are a=6 and b=6
The area of my tent is 36
The Radius of half of the circle is
\sqrt{\frac{6}{2}^2+\frac{6}{2}^2} = \sqrt{ \frac{36}{4} - \frac{36}{4}} = \sqrt{ \frac{72}{4}}
The formula of the circle in term z is
r= \sqrt{\frac{72}{4} -z^2 }
My scaling factor is
s=\frac{\sqrt{72/4-z^2}}{\sqrt{72/4}}
My area formula is:
A(z)= 36 \times \frac{72/4 - z^2}{72/4}
Thus my volume is
36 \int_0^{\sqrt{72/3}} \frac{72/4 - z^2}{72/4} \, dz = \frac{36 \sqrt{72}}{3}
mwilli23
The dimensions of my tent are a=7 and b=5
The area of the base of my tent is 7 *5 = 35
The radius of the half circle is:
\sqrt{\frac{7}{2}^2+\frac{5}{2}^2}=\sqrt{\frac{65}{4}}
The formula for my half circle is:
r=\sqrt{\frac{65}{4}-z^2}
My scaling factor is:
s=\frac{\sqrt{\frac{65}{4}-z^2}}{\sqrt{\frac{65}{4}}}
Volume becomes:
35\int_0^{\sqrt{\frac{65}{4}}}\frac{{\frac{65}{4}}-z^2}{{\frac{65}{4}}}dz ≈94.06