Question

Find the exact area of the surface obtained by rotating the curve about the y -axis. x=√a2−y2, 0≤y≤a/7

Answer 1

Answer:

The exact area of the surface obtained by rotating the curve about the y -axis=$\frac{2\pi a^2}{7}$

Step-by-step explanation:

We are given that the curve

$x=\sqrt{a^2-y^2},0\leq y\leq a/7$

We have to find the exact  area of the surface obtained by rotating the curve about the y -axis.

We know that

Area of the surface obtained by rotating the curve about the y -axis is given by

A=$2\pi\int_{a}^{b}g(y)\sqrt{1+(g'(y))^2}dy$

We have $g(y)=\sqrt{a^2-y^2}$

$g'(y)=-\frac{y}{\sqrt{a^2-y^2}}$

Using the values

a=0,b=a/7

Area of the surface obtained by rotating the curve about the y -axis is given by

A=$2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\sqrt{1+\frac{y^2}{a^2-y^2}}dy$

=$2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\sqrt{\frac{a^2-y^2+y^2}{a^2-y^2}}dy$

=$2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\times \frac{a}{\sqrt{a^2-y^2}}dy$

=$2\pi a \int_{0}^{a/7} dy$

=$2\pi a[y]^{a/7}_{0}$

A=$\frac{2\pi a^2}{7}$

Hence, the  area of the surface obtained by rotating the curve about the y -axis is given by

A=$\frac{2\pi a^2}{7}$