# How do you find the volume of the solid enclosed by the surface z=xsec^2(y) and the planes z=0, x=0,x=2,y=0, and y=π/4?

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To find the volume of the solid enclosed by the surface ( z = x \sec^2(y) ) and the planes ( z = 0 ), ( x = 0 ), ( x = 2 ), ( y = 0 ), and ( y = \frac{\pi}{4} ), you can use a triple integral.

Since the region is bounded by planes, you can set up the integral as follows:

[ V = \int_{0}^{2} \int_{0}^{\frac{\pi}{4}} \int_{0}^{x \sec^2(y)} dz , dy , dx ]

This integral integrates ( z ) from 0 to ( x \sec^2(y) ), ( y ) from 0 to ( \frac{\pi}{4} ), and ( x ) from 0 to 2.

You would first integrate with respect to ( z ), then ( y ), and finally ( x ) to find the volume.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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