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.
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.
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.
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.
$\displaystyle Volume = \pi \int^1_0 (7-4x)^2 - (4-x^2)^2 dx$
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)
$\displaystyle \text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 - \left(\sqrt{4-y}\right)^2\, dy$
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?
i get :
$\displaystyle \pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy - \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$
=$\displaystyle \dfrac{\pi}{16} \left[49y-\frac{14y^2}{2}+ \frac{y^3}{3}\right]^7_3 - \pi\left[4y - \frac{y^2}{2} \right]^4_3$
$\displaystyle \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4) - \frac{4^2}{2}- 4(3) - \frac{3^2}{2} \right]$
=$\displaystyle \dfrac{\pi}{16} \left[\frac{343}{3} - 147 + 63 - 9\right] - \pi\left[8 - \frac{15}{2}\right]$
=$\displaystyle \dfrac{\pi}{16} \left[\frac{64}{3}\right] - \pi\left[\frac{1}{2}\right]$
=$\displaystyle \frac{4}{3}\pi - \frac{1}{2}\pi$
=$\displaystyle \frac{5}{6} unit^3$