(a) See the attached sketch. Each shell will have a radius <em>y</em> chosen from the interval [2, 4], a height of <em>x</em> = 2/<em>y</em>, and thickness ∆<em>y</em>. For infinitely many shells, we have ∆<em>y</em> converging to 0, and each super-thin shell contributes an infinitesimal volume of
2<em>π</em> (radius)² (height) = 4<em>πy</em>
Then the volume of the solid is obtained by integrating over [2, 4]:

(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - <em>x</em> (this is the distance between a given <em>x</em> value in the orange shaded region to the axis of revolution) and a height of 8 - <em>x</em> ³ (and this is the distance between the line <em>y</em> = 8 and the curve <em>y</em> = <em>x</em> ³). Then each shell has a volume of
2<em>π</em> (9 - <em>x</em>)² (8 - <em>x</em> ³) = 2<em>π</em> (648 - 144<em>x</em> + 8<em>x</em> ² - 81<em>x</em> ³ + 18<em>x</em> ⁴ - <em>x</em> ⁵)
so that the overall volume of the solid would be

I leave the details of integrating to you.
Answer:
4m
Step-by-step explanation:
the ratio to find a square’s area is <em>side length</em> times <em>side length</em>, i.e.
sl * sl = a
2m * 2m = 4m
hope this helps!
<span>THE ANSWER:
</span>112 - 7 = 105
105 <span>÷ 7 = 15
</span>x = 15
You multiply using the multiplication table. Multiplication is one of the four basic operations in arithmetic, along with addition, subtraction, and division. Multiplication can actually be considered repeated addition, and you can solve simple multiplication problems by adding repeatedly. For larger numbers, you'll want to do long multiplication, which breaks the process down into repeated simple multiplication and addition problems. You can also try a shortcut version of long multiplication by splitting the smaller number in the problem into tens and ones, but this works best when the smaller number is between 10 and 19.
14 and the remainder is 3