problem in region bounded by a curve and a straight line

**mastermin346** · October 17th 2010, 06:08 AM

Diagram shows a curve $y = 4 - x^2$ and a straight line $y=7 - 4x$ which is tangent to the curve at the point $(1 , 3)$

a)calculate the area of the shaded region.

**Unknown008** · October 17th 2010, 06:14 AM

There is no diagram...

I guess it's the area between the line and the curve. Use integration and the area of a trapezium.

$\text{Area of trapezium} - \int^1_0 4 - x^2 dx$

$\frac12(7 + 3)(1) - \int^1_0 4 - x^2 dx$

There you are!

EdIT: Ah, that's what I thought the diagram was.

**mastermin346** · October 17th 2010, 06:23 AM

How about the volume of revolution,in term of $\pi$ when the shaded region is rotated $360^\circ$ about the $y-$ axis?

**Unknown008** · October 17th 2010, 06:27 AM

Well, you use the formula:

$Volume = \pi \int^1_0 y^2 dx$

Here, it'll be preferable to use the equation of the straight line itself instead of a shortcut like I did for the trapezium above.

$Volume = \pi \int^1_0 (7-4x)^2 - (4-x^2)^2 dx$

**mastermin346** · October 17th 2010, 06:31 AM

are u sure use the formula $Volume = \pi \int^1_0 y^2 dx$ not
$Volume = \pi \int^1_0 x^2 dy$ ?

because it rotated $360^\circ$ about the y-axis

**Unknown008** · October 17th 2010, 06:34 AM

Oh, I didn't notice you were rotating around the y-axis my bad...

Right, we use the formula:

$\text{Volume} =\displaystyle \pi \int^b_a x^2\, dy$

Since you are rotating through the y-axis, you need the y coordinates.

So, we get:

y = 7 - 4x
x = (7 - y)/4

y = 4 - x^2
x = \sqrt(4 - y)

$\text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 - \left(\sqrt{4-y}\right)^2\, dy$

**mastermin346** · October 17th 2010, 06:46 AM

sir,at the curve $y = 4-x^2$

why we use $\pi \int^7_3 \left(\sqrt{4-y}\right)^2 dy$ ?

not use $\pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$ ?

**Unknown008** · October 17th 2010, 06:50 AM

You're right again >_<

I must be tired... sorry for that

**mastermin346** · October 17th 2010, 06:51 AM

oh..its ok sir..u need to rest..thanks for the help sir!

**mastermin346** · October 17th 2010, 07:14 AM

hi..check this:

$\text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2-\pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$

$=\displaystyle \pi \int^7_3 \left(\dfrac{49-y^2}{16}\right) - \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$

= $\pi \left[\frac{49y - y^3}{16(3)}\right]^7_3 - \pi\left[4y - y^2\right]^4_3$

= $\left [\frac{49(7) - 7^3}{48} - \frac{49(3) - 3^3}{48} \right] - \pi\left[4(4) - 4^2 - 4(3) - 3^2\right]$

= $\pi \left( -\frac{5}{2} \right) - \pi ( -21)$

= $\frac{37}{2} \pi unit^3$

am i right sir?

**Unknown008** · October 17th 2010, 07:23 AM

No

$(7-y)^2 = 49 - 14y + y^2$

And $\int 4-y\ dy = 4y - \dfrac{y^2}{2} + c$

**mastermin346** · October 17th 2010, 07:30 AM

urm..

why you different $(7-y)^2 = 49 - 14y + y^2$ ?

and how to integrate $\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2$

**Unknown008** · October 17th 2010, 07:36 AM

$\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 = \pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy= \dfrac{\pi}{16} \int^7_3 49-14y+y^2 \ dy$

**mastermin346** · October 17th 2010, 07:38 AM

ok thanks..i try do it now..later,,can u check my answer?

**mastermin346** · October 17th 2010, 08:00 AM

i get :

$\pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy - \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$

= $\dfrac{\pi}{16} \left[49y-\frac{14y^2}{2}+ \frac{y^3}{3}\right]^7_3 - \pi\left[4y - \frac{y^2}{2} \right]^4_3$

$\dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) - \frac{3^2}{2} \right]$

= $\dfrac{\pi}{16} \left[\frac{343}{3} - 147 + 63 - 9\right] - \pi\left[8 - \frac{15}{2}\right]$

= $\dfrac{\pi}{16} \left[\frac{64}{3}\right] - \pi\left[\frac{1}{2}\right]$

= $\frac{4}{3}\pi - \frac{1}{2}\pi$

= $\frac{5}{6} unit^3$

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