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Find the area of the region in the first quadrant enclosed by the y-axis,the line `y=x`and the circle `x^2+y^2=32 ,`using integration. |
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Answer» `r=sqrt32=4sqrt2` `y=r=4sqrt2` `2y^2=32` `y^2=16` `y=4` `area=int_0^4sqrt(32-x^2)-xdx` `=int_0^4sqrt((4R)^2-x^2)dx-int_0^4xdx` `=(4pi-8)-[x^2/2]_0^4` `=4pi-8-[(16-0)/2]` `=4pi-8-8` `=4pi-16`. |
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