Answer:
77/24
The answer for this question is 77/24
The region in question is the set
or equivalently,
![R = \left\{ (x, y) : 0 \le y \le 1 \text{ and } 0 \le x \le \sqrt[3]{y} \right\}](https://tex.z-dn.net/?f=R%20%3D%20%5Cleft%5C%7B%20%28x%2C%20y%29%20%3A%200%20%5Cle%20y%20%5Cle%201%20%5Ctext%7B%20and%20%7D%200%20%5Cle%20x%20%5Cle%20%5Csqrt%5B3%5D%7By%7D%20%5Cright%5C%7D)
Cross sections are taken perpendicular to the y-axis, which means each section has a base length equal to the horizontal distance between the curve y = x³ and the line x = 0 (the y-axis). This horizontal distance is given by
y = x³ ⇒ x = ∛y
so that each triangular cross section has side length ∛y.
The area of an equilateral triangle with side length s is √3/4 s², so each cross section contributes an infinitesimal area of √3/4 ∛(y²).
Then the volume of this solid is
![\displaystyle \frac{\sqrt3}4 \int_0^1 \sqrt[3]{y^2} \, dy = \frac{\sqrt3}4 \int_0^1 y^{2/3} \, dy = \frac{\sqrt3}4\cdot\frac35 y^{5/3} \bigg|_0^1 = \boxed{\frac{3\sqrt3}{20}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20%5Csqrt%5B3%5D%7By%5E2%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%20%5Cint_0%5E1%20y%5E%7B2%2F3%7D%20%5C%2C%20dy%20%3D%20%5Cfrac%7B%5Csqrt3%7D4%5Ccdot%5Cfrac35%20y%5E%7B5%2F3%7D%20%5Cbigg%7C_0%5E1%20%3D%20%5Cboxed%7B%5Cfrac%7B3%5Csqrt3%7D%7B20%7D%7D)
I've attached some sketches of the solid with 16 and 64 such cross sections to give an idea of what this solid looks like.
Answer:
7/8
Step-by-step explanation:
turn 1/4 denominator into 8 and by doing that, u multiply by 2.
whatever you do to the bottom, u do to the top. so 1/4= 2/8
5/8+2/8= 7/8
Answer:
The value of p = 120 when q = 5
Step-by-step explanation:
We know that if y varies inversely with x, we get the equation
y ∝ 1/x
y = k/x
k = yx
where 'k' is called the constant of proportionality.
In our case,
If p is inversely proportional to the square of q, then
p ∝ 1/q²
p = k/q²
k = pq²
As p is 30 when q is 10, so
substituting p = 30 and q = 10
k = (30) (10)²
k = (30)(100)
k = 3000
Now,
Determining the value of p when q = 5
Using the equation
p = k/q²
now substituting k = 3000 and q = 5
p = 3000 / (5)²
p = 3000/25
p = 120
Therefore, the value of p = 120 when q = 5
Answer:
4.71
Step-by-step explanation:
3* pi = 9.42
9.42/2 = 4.71
