Subtracting areas when doing inverse trig functions? (Urgent)

[Lols123]

(For those with the essentials textbook, referring to question in 8A, q6 where we need to find the areas for inverse trig functions)
just wondering how we know when the area of an inverse trig function will be the same as the normal trig function and when we have to subtract it from the inverse function

from what i have done, generally the area for cos will be same and tan/sin will require you to subtract?

got a sac on monday, so any help really appreciated

[taiga]

What is your question? sorry to sound rude

[Lols123]

how do you find the area for inverse trig functions

[moekamo]

take question 6a for example,



so let

now if you graph y against x you will see that the original integral is the area under y from x=0 to x=1, or alternatively, it is the area of the rectangle(pi/4 units squared) minus the area with the y axis, which is found from:



so

This is quite a general way of solving these problems. I'd advise against learning a specific rule like for cos its the same and sin/tan you minus, because then when there's something a little bit different you can get into trouble if you don't understand why its like that. Also, by understanding, you can easily derive this in an exam by drawing a quick sketch of the curve and avoid useless rote learning.

[Lols123]

hmm so how can you tell whether you need to use that method (subtracting? :S

oh and also, when doing the volumes, how do you tell what shape it is
like for one of the examples it shows that a triangle represents a cone
would that mean that anytime i see a triangle and the question is what shape is it, the answer is a cone?

[moekamo]

With the above example, the integral wants us to find an area(effectively) so we draw the graph and use geometry to reason that the integral is the area of a rectangle minus the area with the y axis. In other cases you will see from the graph that the area with the y axis is the same as the area with the x axis, so you wont need to minus here, but by drawing the graph you will easily see why.

For volumes...i guess you just have to look at the graph and imagine it rotating about an axis to get the desired shape, like if you rotate y=x around the x axis, it will be a cone, i dont think there is much more to it, but with practice you will be able to imagine what the volume looks like. And in most cases anyway, knowing the shape wont help as they wont have pre-determined formulas for the volume anyway...

[Lols123]

ohh thanks! i get it now

just a couple more questions:
1) was asking about the shapes question because my teacher likes to ask them. for example, what shape do you think this would be:
(x-2)^2 + y^2 =1
teacher says its a donut, but i cant really see how that works :/
so yeah...but probs not on the sac anyway

2) for the equation: y = x^2 + 4
would the volume rotated around y be = pi x (antidiff of (y-4)) from 4 to 8?

[brightsky]

1) circle --> remember general circle formula
2) yep, although i don't know where the 8 came from..

[Lols123]

ohh okay, so semi circle = sphere?
circle = donut?

oh sorry, the domain was from 0 to 2
just wondering as well, would you be able to use that volume to find the volume rotated around x by
subtracting it from the integral of   pi x (antidiff ( 2^2)) from 0 - 8
i.e. by using the line x = 2

[brightsky]

depends which axis you rotate it around. x-axis = sphere, y-axis = "donut". all this can be derived from visualisation on a cartesian plane; try not to memorise these things.

soz..not really sure what you're asking for the second question...

[moekamo]

for 1, it depends what axis your rotating about, if its the x axis, then yes its a sphere, however if you rotate about the y axis you will get a donut or 'torus'.

and i dont quite understand the next question, do you want the volume of x^2+4 being rotated about the x axis?

EDIT: beaten

[Lols123]

ah k, kind of hard to visualise though :/

yeah just wondering if you can find the volume rotatated around the x-axis but by using Vy instead of using the Vx formula