You visited us 1 times! Enjoying our articles? Unlock Full Access!

Double Ordinate of Ellipse

The area of t...

Question

The area of the region bounded by $y = (x - 4)^{2}, y = 16 - x^{2}$ and the x axis,is

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

32

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{64}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

64

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{64}{3}$
$y_{1} = x^{2} + 16 - 8 x$
$y_{2} = 16 - x^{2}$
$16 - x^{2} = x^{2} = 16 - 8 x$
$\Rightarrow 2 x^{2} - 8 x = 0$
$x^{2} - 4 = 0$
$x = 0, 4$
$4 \int 0 (y_{1} - y_{2}) d x$
$\Rightarrow \int 0 (2 x^{2} - 8 x) d x$
$\Rightarrow {[\frac{{2 x}^{3}}{3} - {4 x}^{2}]}_{0}^{4}$
$∣ ∣ ∣ \frac{128}{3} - 64 ∣ ∣ ∣ = \frac{64}{3}$

Suggest Corrections

Similar questions

x^{2} = 16 y, y = 1, y = 4

y

8

x -

y -

y = [\frac{x^{2}}{64} + 2], y = x - 1

x = 0

x -

[.]

y = x^{3}

x = 2

x = 4

\frac{8 a^{3}}{3}

\frac{16 a^{2}}{3}

\frac{32 a^{2}}{3}

\frac{64 a^{2}}{3}

Join BYJU'S Learning Program