# Thread: problem in region bounded by a curve and a straight line

1. ## problem in region bounded by a curve and a straight line

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

a)calculate the area of the shaded region.

2. There is no diagram...

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

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

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

There you are!

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

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

4. Well, you use the formula:

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

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.

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

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

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

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

Right, we use the formula:

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

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)

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

7. sir,at the curve $\displaystyle y = 4-x^2$

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

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

8. You're right again >_<

I must be tired... sorry for that

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

10. hi..check this:

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

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

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

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

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

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

am i right sir?

11. No

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

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

12. urm..

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

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

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

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

15. i get :

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

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

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

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

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

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

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

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