Answer:
The exact area of the surface obtained by rotating the curve about the y -axis=[tex]\frac{2\pi a^2}{7}[/tex]
Step-by-step explanation:
We are given that the curve
[tex]x=\sqrt{a^2-y^2},0\leq y\leq a/7[/tex]
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=[tex]2\pi\int_{a}^{b}g(y)\sqrt{1+(g'(y))^2}dy[/tex]
We have [tex]g(y)=\sqrt{a^2-y^2}[/tex]
[tex]g'(y)=-\frac{y}{\sqrt{a^2-y^2}}[/tex]
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=[tex]2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\sqrt{1+\frac{y^2}{a^2-y^2}}dy[/tex]
=[tex]2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\sqrt{\frac{a^2-y^2+y^2}{a^2-y^2}}dy[/tex]
=[tex]2\pi \int_{0}^{a/7}\sqrt{a^2-y^2}\times \frac{a}{\sqrt{a^2-y^2}}dy[/tex]
=[tex]2\pi a \int_{0}^{a/7} dy[/tex]
=[tex]2\pi a[y]^{a/7}_{0}[/tex]
A=[tex]\frac{2\pi a^2}{7}[/tex]
Hence, the area of the surface obtained by rotating the curve about the y -axis is given by
A=[tex]\frac{2\pi a^2}{7}[/tex]
