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Finding the Area of a Shaded Region

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the total area of the shaded region

    yrMcG.png

    2. Relevant equations

    The Fundamental Theorem of Calculus: [itex]\int[/itex][itex]^{b}_{a}[/itex] f(x) dx = F(b) - F(a)

    3. The attempt at a solution

    I don't seem to have a clue about how to approach this one. Though for previous total area problems, I had to take the absolute value of specific intervals and then add them to obtain the area, or subtract the area under the curve by forming a rectangle using the boundaries given, and subtracting the area of the curve from the area of the rectangle.

    Technically I could make a rectangle in this problem, but do I have to? I don't really have any examples in my textbook that point me towards any strategy here.
     
  2. jcsd
  3. Apr 22, 2012 #2
    Technically the area would just be the area of the rectangle because the area under the x-axis has to equal the area from ∏/4 to ∏/2.

    I think that would be right?
     
  4. Apr 22, 2012 #3
    Ah you're right! That bottom portion should fit into the top portion to make a good rectangle :O

    Is there some way to calculate that though?
     
  5. Apr 22, 2012 #4
    Think basic ;)

    Area of a rectangle = Base x Height
     
  6. Apr 22, 2012 #5
    Sorry, haha. I mean, using antiderivatives considering this chapter focuses on using those to calculate areas. I'm aware that it's much simpler to just form a rectangle of [itex]\frac{\pi}{2}[/itex] base and 4[itex]\sqrt{2}[/itex] height, and multiply those values to get the area, but I'm curious as to how one obtains it through the use of antiderivatives.
     
    Last edited: Apr 22, 2012
  7. Apr 22, 2012 #6
    You would calculate the integral: ##\displaystyle \int_{\pi/4}^{3 \pi/4} 4\sqrt{2} - 4\csc{\theta}\cot{\theta} \ d\theta##

    This is because the ##4\sqrt{2}## term calculates the area of the box. By subtracting the second part, you take away from the first part of the box, and then you ADD the second part of the integral of the curve. Normally the area is negative for the part of the curve below the x-axis, but when you subtract it becomes positive.
     
  8. Apr 22, 2012 #7
    Ah alright! That worked for me.

    I was on the right track before, but I went ahead and separated the integral into two parts so that I'd have one integral from [itex]\frac{\pi}{4}[/itex] to [itex]\frac{\pi}{2}[/itex] and another from [itex]\frac{\pi}{2}[/itex] to [itex]\frac{3\pi}{4}[/itex]. I see how that works now though.

    Thank you for the help!
     
    Last edited: Apr 22, 2012
  9. Apr 22, 2012 #8

    sharks

    User Avatar
    Gold Member

    Well, you could solve this using double integral, which i doubt you've covered yet in your syllabus. But i don't think that's what you're being required to do in this problem. You should just solve it as suggested above.
     
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