Площадь и объем тела вращения

[CrypticMath]

Привет,

I need some help with these.
Спасибо за помощь!

Consider the region bounded by the following curves $y=14-x^2, y=x^2-4, \text{and} x=0, \text{for}\ x \geq 0$:

1.) set up an integral expression that would give the area of the region of $y$ as a function of $x$:

x^2-4=14-x^2$
$2x^2-18=0$
$2(x-3)(x+3)=0$
$x=-3; x=3$
$\text{Integral:}\ \int_{-3}^{3} |\left(x^2-9\right)| dx$

2.) set up an integral expression that would give the area of the region of $x$ as a function of $y$:

$\int_{-4}^{14} |\sqrt{14-y}| dy$

3.) compute the area of the region by evaluating one of the expressions in (1.) or (2.):

$\int_{-3}^{3} |\left(x^2-9\right)| dx$
$=\dfrac{x\left(x^2-27\right)}{3}+\text{C}$
$\int_{-3}^{3} \left(x^2-9\right) dx$

$|-18-18| = 36\ \text{square units}$


4.) set up an integral expression that would give the volume of the solid created by rotating the region about [TEX]y=20[/TEX], using the disk/washer method:

$\pi \int_0^{20} \left(2x^2\right)^2-18x^2 dx$


5.) find the volume of the solid by evaluating the integral/integrals in (4.):

$\pi \int_0^{20} \left(2x^2\right)^2-18x^2 dx$
$=\dfrac{4x\left(x^4-405\right)}{5}+ \text{C}$
$\int_0^{20} \left(2x^2\right)^2-18x^2 dx = 2553520\pi\ \text{cube units}$

6.) set up an integral expression that would give the volume of the solid created by rotating the region about $x=4$, using the disk/washer method:

$\pi \int_0^4 \left(\sqrt{14-y}\right)^2 - \left(\sqrt{4+y}\right)^2 dy$

7.) compute the volume of the solid by evaluating the integral or integrals in (6.):

$\pi \int_0^4 \left(\sqrt{14-y}\right)^2 - \left(\sqrt{4+y}\right)^2 dy$
$\int \left(\sqrt{14-y}\right)^2 - \left(\sqrt{4+y}\right)^2 dy\ =\ -\left(y-10)y\right)\,+\,\text{C}$

$\displaystyle \lim_{y \to 4^{-}} -\left(y-10)y\right)\,=\,24$

$\displaystyle \lim_{y \to 0^{+}} -\left(y-10)y\right)\,=\,0$

$24\pi\ \text{cube units}$

[Otta]

CrypticMath в сообщении #846737 писал(а):

$y=14-x^2, y=x^2-4, \text{and} x=0, \text{for}\ x \geq 0$:

1) Неверно. Не все условия учтены.
2) Просто неверно. Попробуйте рисовать графики.
3) Нет смысла обсуждать (см. выше).
4) Нет.
... Все, достаточно.