#### Question

Find the area of elipse `x^2/a^2+y^2/b^2=1`

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#### Solution

equation of elipse `x^2/a^2+y^2/b^2=1`

Clearly the area of elipse is 4 times the area of region OPQD as show in the figure.For the region.Limits of integration are x=0 and x=a

From the elipse

`y^2/b^2=1-x^2/a^2=(a^2-x^2)/a^2`

`therefore y^2=b^2/a^2(a^2-x^2)`

`y=+-b/a(a^2-x^2)^(1/2)`

`y=b/a(a^2-x^2)^(1/2)`

We know

`A=4int_0^1ydx`

`=int_0^ab/a(a^2-x^2)^(1/2)dx`

`=(4b)/a[x/2(a^2-x^2)^(1/2)+a^2/2sin^(-1)(x/a)]_0^a`

`=(4b)/a[a^2/2 pi/2-0]`

`=ab pi `

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#### APPEARS IN

Solution for question: Find the area of elipse x^2/a^2+y^2/b^2=1 concept: null - Applications of Definite Integrals. For the courses HSC Commerce, HSC Commerce (Marketing and Salesmanship)