Calculus: Area Volume

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May 2nd 2009, 12:45 PM
Kayla_N

Calculus: Area Volume

1. http://webwork1.math.utah.edu/webwor...rob5image1.png
Find the area enclosed between http://webwork1.math.utah.edu/webwor...98387e6711.png and http://webwork1.math.utah.edu/webwor...d9d094b661.png, on the interval from http://webwork1.math.utah.edu/webwor...d394853bc1.png to http://webwork1.math.utah.edu/webwor...a1091cf1a1.png.
Area =????

2. Find the area of the region enclosed between http://webwork1.math.utah.edu/webwor...edfca60801.png and http://webwork1.math.utah.edu/webwor...930be57411.png from http://webwork1.math.utah.edu/webwor...c67fef88e1.png to http://webwork1.math.utah.edu/webwor...7b9d151061.png. Hint: Notice that this region consists of two parts.
May 2nd 2009, 12:58 PM
Spec

If $f(x) \geq g(x)$ on the intervall I, then the area is $\int_I (f(x)-g(x))dx$

The procedure is the same for the second problem, only that you you might need to split the integral into smaller intervals and switch the functions around (try to graph the functions to get a better overview).
May 2nd 2009, 01:09 PM
Pinkk

Notice how $f(x)>g(x)$ on $[-3,7]$ . So the area is evaluated as:

$\int_{-3}^{7}\frac{x^{2}}{4}+10-x\,\,dx$

Simply evaluate the integral from there.

For the next problem, find where the two graphs intersect and knowing how sine and cosine curves are on the given interval (which is $[0,\frac{7\pi}{10}]$ ), you'll know that to the left of the intersection $3cos(x)>2sin(x)$ and to the right of it $2sin(x)>3cos(x)$

$3cos(x)=2sin(x)$
$tan(x)=\frac{3}{2}$
$x=tan^{-1}(\frac{3}{2})$

So the two intergrals needed are:

$\int_{0}^{tan^{-1}(\frac{3}{2})}3cos(x)-2sin(x)\,\,dx + \int_{tan^{-1}(\frac{3}{2})}^{\frac{7\pi}{10}}2sin(x)-3cos(x)\,\,dx$
May 2nd 2009, 01:22 PM
Kayla_N

So 1st in i plug everyting in and got 0. second one is 3.381389746??
May 2nd 2009, 01:29 PM
Pinkk

The first one is definitely wrong. You can tell by the graph itself that it has to have an area greater than zero. Check your work again.
May 2nd 2009, 01:29 PM
Spec

According to your graph the first area is obviously not zero.

EDIT: Beat me by seconds. ^_^
May 2nd 2009, 01:51 PM
Kayla_N

So for the 1st one i plug in 7 and - the result when i plug in neg 3. I got 0

ANyways thanks for your help..I give up. :)
May 2nd 2009, 02:07 PM
Spec

$\int_{-3}^{7}\left(\frac{x^{2}}{4}+10-x\,\right)\,dx=\left[ \frac{x^3}{12} +10x - \frac{x^2}{2}\right]_{-3}^7$ $=\left[ \frac{x^3+120x-6x^2}{12}\right]_{-3}^7=\frac{889-(-441)}{12}=\frac{1330}{12}=\frac{665}{6}$
May 2nd 2009, 02:24 PM
Kayla_N

Oh i c i forgot the power rule..Thanks. for the second i just plug it in right?
May 2nd 2009, 02:55 PM
Pinkk

What do you exactly mean by plug in? Evaluate the integrals and then evaluate them with the respective limits of integration.
May 2nd 2009, 03:36 PM
Kayla_N

Thanks i got it 3.959622071.

Quote:

Originally Posted by Pinkk

What do you exactly mean by plug in? Evaluate the integrals and then evaluate them with the respective limits of integration.