$\displaystyle \int_{-3}^{0} 1+ \sqrt{9-x^2} \, dx$

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- Dec 31st 2011, 12:06 PMmathkEvaluate the integral by interpreting it in terms of area
$\displaystyle \int_{-3}^{0} 1+ \sqrt{9-x^2} \, dx$

- Dec 31st 2011, 12:50 PMPlatoRe: Evaluate the integral by interpreting it in terms of area
- Dec 31st 2011, 01:00 PMPlatoRe: Evaluate the integral by interpreting it in terms of area
- Dec 31st 2011, 01:02 PMmathkRe: Evaluate the integral by interpreting it in terms of area
ok What is OP and how do you know its a rectangle and a circle?

- Dec 31st 2011, 01:09 PMPlatoRe: Evaluate the integral by interpreting it in terms of area
- Dec 31st 2011, 01:11 PMmathkRe: Evaluate the integral by interpreting it in terms of area
- Dec 31st 2011, 03:07 PMHallsofIvyRe: Evaluate the integral by interpreting it in terms of area
I have no idea what "umm ok" means!

The problem asked you to evaluate this by interpreting it as an area so you should have thought about the**geometry**determined by these functions' graphs. The integral can be interpreted as "the area under the graph". Your integrand was $\displaystyle 1+ \sqrt{9- x^2$.

You should, perhaps from pre-calclulus, perhaps from Cartesian geometry, recognise that the graph of y= 1 is a horizontal straight line and, from x= -3 to 0, from y= 0 to y= 1, is a rectangle with length 0-(-3)= 3 and height 1. What is the area of that rectangle?

You should also be able to recognixze that $\displaystyle y= \sqrt{9 -x^2}$ is the part of the**circle**$\displaystyle x^2+ y^2= 9$ above the x-axis (because the square root is positive). That circle has center (0, 0) and radius 3 and so goes, along the x-axis, from -3 to 3. Going from -3 to 0 gives you half of that circle and taking y positive gives you half of**that**- one quarter of the circle. What is that area? - Dec 31st 2011, 04:28 PMProve ItRe: Evaluate the integral by interpreting it in terms of area
The geometric interpretation is the easiest method, as has been pointed out to you. But if you were going to use the substitution, let $\displaystyle \displaystyle \begin{align*} x = 3\sin{\theta} \implies dx = 3\cos{\theta}\,d\theta \end{align*}$, and note that when $\displaystyle \displaystyle \begin{align*} x = -3, \theta = -\frac{\pi}{2} \end{align*}$ and when $\displaystyle \displaystyle \begin{align*} x = 0, \theta = 0 \end{align*}$, then the integral becomes...

$\displaystyle \displaystyle \begin{align*} \int_{-3}^0{ 1 + \sqrt{ 9 - x^2 } \, dx } &= \int_{-3}^0{1\,dx} + \int_{-3}^0{\sqrt{9-x^2}\,dx} \\ &= \left[ x \right]_{-3}^0 + \int_{ -\frac{ \pi }{ 2 } }^0{ \sqrt{ 9 - \left( 3\sin{\theta} \right)^2 } \, 3\cos{\theta} \,d\theta } \\ &= 0 - (-3) + 3\int_{-\frac{\pi}{2}}^0{ \sqrt{9 - 9\sin^2{\theta}} \, \cos{\theta}\,d\theta } \\ &= 3 + 3\int_{-\frac{\pi}{2}}^0{\sqrt{9\left( 1 - \sin^2{\theta} \right)} \, \cos{\theta}\,d\theta} \\ &= 3 + 3 \int_{-\frac{\pi}{2}}^0{ \sqrt{ 9\cos^2{\theta} } \,\cos{\theta}\,d\theta} \\ &= 3 + 3\int_{-\frac{\pi}{2}}^0{ 3 \cos{\theta} \cos{\theta}\,d\theta} \\ &= 3 + 9\int_{-\frac{\pi}{2}}^0{\cos^2{\theta}\,d\theta} \\ &= 3 + 9\int_{-\frac{\pi}{2}}^0{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= 3 + 9\left[ \frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} \right]_{-\frac{\pi}{2}}^0 \end{align*}$

$\displaystyle \displaystyle \begin{align*} &= 3 + 9\left\{ \left[ \frac{1}{2}\cdot 0 + \frac{1}{4}\sin{(2\cdot 0)} \right] - \left[ \frac{1}{2}\left(-\frac{\pi}{2}\right) + \frac{1}{4}\sin{ \left[2 \left( -\frac{\pi}{2} \right) \right] } \right] \right\} \\ &= 3 + \frac{9\pi}{4} \end{align*}$

which is the same as the area of the rectangle of length 3 units and width 1 unit plus the area of the quarter circle of radius 3 units, as you were advised :) - Dec 31st 2011, 05:35 PMPlatoRe: Evaluate the integral by interpreting it in terms of area
@Prove It, Thank you for your contribution to this question.

However, I must say that I think it violates the sprit of the question.

I think that your approach is a least twenty years out of date.

That is, I think the author of the question is spot on to the current state in mathematics education. I have a firmly held thought that within twenty years no mainstream calculus text will contain a chapter on “techniques of integration”. The wide availability of computer algebra systems have made those topics obsolete for the most part. (Of course they will be great topics for Mathematics Education courses and courses in the History of Mathematics.)

Look at this link to see what I mean. - Dec 31st 2011, 08:04 PMProve ItRe: Evaluate the integral by interpreting it in terms of area
Violates the spirit of the question, perhaps.

Obsolete, hardly. There are many integrals that would require trigonometric/hyperbolic substitution that can't be solved using a geometric interpretation. And the use of a CAS in those situations is just plain lazy and irresponsible. How would they be programmed in future if everyone thought the method of solution was "obsolete"?

Besides, my post was in response to the original poster asking for clarification when someone else suggested using this method. I pointed out that the geometric interpretation is the quicker and better method. - Jan 1st 2012, 12:59 AMCaptainBlackRe: Evaluate the integral by interpreting it in terms of area
I think in what passes for the real world they have been out of date for far longer. In fields that actually use such integrals it has been virtually universal practice to look them up in tables since before I started work outside of a university in the late 70's. The tables by Gradshteyn & Ryzhik have been the standard since the mid 60's, when they replaced tables going back at least as far as the 1820's.

CB