# Finding area and volume of bounded region via integration

1. Apr 7, 2014

### Sociomath

Hi,

I just need these solutions checked.

Consider the region bounded by the following curves $y=x-3, y=5-x, \text{and}\ y=3$:

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

$y = x-3 = 5-x$
$x + x - 3 - 5 = 0$
$2x-8=0$
$x = 4$
$\displaystyle \int_0^3 |\left(2x-8\right)|\ dx$

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

$\displaystyle \int_0^5 |\left(5-y\right)|$

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

$\displaystyle \int_0^3 |\left(2x-8\right)|\ dx$
$\left(x-8\right)+\ \text{C}$
$\displaystyle = \lim_{x \to 0^{+}} ((x-8)x)=0$
$\displaystyle = \lim_{x \to 3^{-}} ((x-8)x)=-15$
$\displaystyle |-15-0| = 15$

4.) set up an integral expression that would give the volume of the solid created by rotating the region about y = 1, using the disk/washer method:

$\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx$

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

$\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx$
$=\ -\dfrac{29}{6}\pi$
$\approx 15\ \text{cubic units}$

6.) set up an integral expression that would give the volume of the solid created by rotating the region about y = -2, using the disk/washer method:

$\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx$

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

$\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx$
$=\ -\dfrac{27}{2}\pi$
$\approx 42\ \text{cubic units}$

2. Apr 7, 2014

### LCKurtz

I would suggest you start over and begin by drawing a picture of the region. You are going to need two integrals using either a dx or dy variable. Also, what bounds the region on the left?

3. Apr 7, 2014

### az_lender

You were correct in identifying x=4 as an intersection point, and y = 1 there; but the other two intersection points must occur where y = 3, so x is certainly NOT zero or 3 ! So there is no reason to integrate from 0 to 3 with respect to x.

4. Apr 13, 2014

### Sociomath

Thanks LCKurtz and az_lender!

1. $\displaystyle \int_2^4 \left(3-\left(5-x\right)\right)dx\,+\,\displaystyle \int_4^6\left(\left(3-\left(x-3\right)\right)\right)dx$
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2. $\displaystyle \int_1^3 ((y+3-(5-y))dy\,=\,\int_1^3 (2y-2))dy$
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3. Using (1.) from above:
$\displaystyle \int_2^4 \left(3-\left(5-x\right)\right)dx\,+\,\displaystyle \int_4^6\left(\left(3-\left(x-3\right)\right)\right)dx$
$=\,\dfrac{x^2}{2}\,+\,C$
$\displaystyle\lim_{x\to 2^{+}} \left(\dfrac{x^2}{2}\right)\,=\,2$
$\displaystyle\lim_{x\to 4^{-}} \left(\dfrac{x^2}{2}\right)\,=\,8$

$\text{From the above}\,=\,8-2\,=\,6$
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4. $\displaystyle \int_2^4 \pi ((2)^2-((3-(5-x)))^2 dx\,+\,\int_4^6 \pi ((2)^2\,-\,(3-(x-3)))^2 dx$
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5. From (4.) above:
$\dfrac{16\left(1-3\pi\right)\pi}{3}$
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6. $\displaystyle \int_1^3 \pi \left(6-\left(5-y\right)\right)^2\,-\,\left(6\,-\,\left(y\,+\,3\right)\right)^2dy\ \left(6-\left(5-y\right)\right)^2\,-\,\displaystyle \int \left(6\,-\,\left(y\,+\,3\right)\right)^2dy\ \,=\,4\left(y\,-\,2\right)y\,+\,\dfrac{28}{3}\,+\,C$
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7. From (6.) above:
$\displaystyle\lim_{y\to 1^{+}} (4(y-2))y\,+\,\dfrac{28}{3})\,=\,\dfrac{16\pi}{3}$
$\displaystyle\lim_{y\to 3^{-}} (4(y-2))y\,+\,\dfrac{28}{3})\,=\,\dfrac{64\pi}{3}$

$=\,\dfrac{64\pi}{3}\,-\,\dfrac{16\pi}{3}\,=\,16\pi$

Last edited: Apr 13, 2014
5. Apr 13, 2014

### LCKurtz

Those look correct for the triangular area.