Shown below is an ellipse whose equation is `frac{x^2}{81}` + `frac{y^2}{36}=1`

Find the area of the shaded region in terms of `pi`. Show your steps.

Find the area of the shaded region in the figure given below [Take pi=3.14 ]

Find the area under the shaded region for curve y=3x^(2) .

Find the area of the ellipse x^(2)/64 + y^(2)/36 = 1 .

Consider the family of circles x^2 + y^2= r^2 2 lt r lt 5 . If in the first quadrant, the common tangent to a circle of this family and the ellipse frac{x^(2)}{25} +frac{y^(2)}{4} =1 meets the axes at A and B then find the equation of the locus of middle point of AB.

Find the area of ellipse x^(2)/1 + y^(2)/4 = 1.

Find the area of the ellipse x^2 + 9y^2 = 36 using integration

Figure shows the curve y=x^2 . Find the area of the shaded part between x=0 and x=6 .

Find the radius of the sphere whose surface area is 36pi cm^(2) .

Find the area of the region enclosed by the given curves y=cos x,y=1-(2x)/(pi)

Let frac{x^2}{a^2}+ frac{y^2}{b^2}=1 , a gt b be an ellipse, whose eccentricity is frac{1}{sqrt 2} and the length of the latus rectum is sqrt (14) . Then the square of the eccentricity of frac{x^2}{a^2}- frac{y^2}{b^2}=1 is