Hey everyone. This question got posted in the 2U question forum, it shows the worst a volume of revolution question could possibly get (if BOSTES feels really mean, anyway). Check out the working! It was new for me too, so if anyone does spot an error or has some feedback, give a holler!
i need integration explanations please
1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4
Hey hey! Thanks to cosine for the awesome explanation for the first half of that explanation. I'll tag in for the second half
This one is tricky! More like what you would see in an Extension One Paper. I'll also mention that this question wouldn't be asked in the HSC, as it isn't specific enough. Specifically, we don't know whether the question wishes us to find the volume enclosed by the parabola, and the lines y=1 and y=4, or otherwise. A question specifying a single arm is a possibility though. But since I'm a sucker for punishment, I'm going to assume it wants the area between the arms, as it is also the most nasty route. I've actually never done a question like this before, so here goes!
The first thing we need to do is rearrange the given function to make x the subject:
 }^{ 2 }\\ x+5=\pm \sqrt { y } \\ x=-5\pm \sqrt { y } )
The plus/minus represents the two arms of the parabola. If they want the area enclosed by the curve and the two lines, we need to use both arms. The arm furthest from the y axis will give us our first volume, this happens to be the minus. The other will give us a smaller volume, as it is closer to the y axis. The larger volume, minus the smaller volume, gives us the volume enclosed by the parabola. Hopefully that makes sense! If you draw a diagram, you will see that the arm represented by
is further from the y axis than the other. To find the area enclosed, we take this larger area, minus the smaller area under the closer arm, denoted by
.
Actually, never mind, here is a diagram! Larger volume in red, smaller in blue! The area we want is red-blue, simple!
I'll denote the arm which will give the larger volume as
, and the smaller as
.
The volume can be derived as follows.
 } dy\\ \\ \quad =\pi \int _{ 1 }^{ 4 }{ \left( \left( 25+10\sqrt { y } +y \right) -\left( 25-10\sqrt { y } +y \right) \right) } dy\\ \\ \quad =\pi \int _{ 1 }^{ 4 }{ \left( 20\sqrt { y } \right) } dy\\ \\ \quad =20\pi \left[ \frac { 2 }{ 3 } \right { \left { y }^{ \frac { 3 }{ 2 } } \right] }_{ 1 }^{ 4 }\\ \\ \quad =\frac { 40\pi }{ 3 } \left[ 8-1 \right] \\ \\ \quad =\frac { 280\pi }{ 3 } { u }^{ 3 } )
I'll stress that, unless this is a 4 unit thing (which I am not familiar with), this will almost definitely NOT be asked in the HSC. If it was, it would lead you through the process, rather than hang you out to dry like that. And this would be in a 3 unit paper, it would definitely never go near a 2 unit paper, simply too intricate. Hopefully that answers the question, and hopefully it is correct! If anyone spots an error please shout out and I'll make amendments, or open the floor to someone more knowledgeable than I, this was new for me too!
PS- If you wanted to do it with just one specific arm, you would simply find the volume for that arm only. Much easier.
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